Limits and Derivatives

1. CONCEPT OF LIMITS

1. INTRODUCTION

Consider the function $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2–1}{\mathrm{x}–1}$. Clearly f(x) is not defined at x = 1. At x = 1, $\mathrm{f}\left(\mathrm{x}\right)=\frac{1–1}{1–1}=\frac{0}{0}$, which is meaningless.

X	.9	.99	.999	.9999	.99999
f(x)	1.9	1.99	1.999	1.999	1.99999

From the above table it is clear that as x approaches to 1 i.e. x $\to$ 1 from the left hand side (means
x approaches 1 from the values less than 1) f(x) approaches to 2 i.e. f(x) $\to$ 2. The number 2 is called the left limit of f(x) and in symbol we shall write $\underset{\mathrm{x}\to 1^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=2$

Again let us study the behaviour of f(x) where x approaches towards 1 from the right-hand side.

X	1.1	1.01	1.001	1.0001	1.00001
f(x)	2.1	2.01	2.001	2.0001	2.00001

It is clear from the table that as x approaches to 2 i.e. x $\to$ 2, from the right-hand side (means x
approaches 1 from the values greater than 1) f(x) approaches to 2 i.e. f(x) $\to$ 2. Here 2 is called the
right-hand limit of f(x) and in symbol we will write $\underset{\mathrm{x}\to 1^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=2$

Thus we see that f(x) is not defined at x = 1 but its left-hand limit and right-hand limit as x $\to$ 1 exist and are equal. When $\underset{\mathrm{x}\to 1^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right) \mathrm{and} \underset{\mathrm{x}\to 1^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ are equal we say $\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ exist and is equal to 2.

2. MEANING OF ‘ x $\to$ a ’

Let x be a variable and ‘a’ be a constant. If x assumes values nearer and nearer to ‘a’ but x is strictly
smaller than ‘a’ then this statement is mathematically written as $\mathrm{x}\to {\mathrm{a}}^{–}$.

Similarly, $\mathrm{x}\to {\mathrm{a}}^{+}$, implies that x assumes values nearer and nearer to ‘a’ but x is strictly greater
than ‘a’.
In general by ‘x tends to a’ we mean that
(i) x ≠ a
(ii) x assumes values nearer and nearer to ‘a’ and
(iii) we are not specifying the manner in which x should approach to ‘a’.

x may approach to ‘a’ from left or right as shown in figure.

If ‘x’ approach to ‘a’ from any point on the right of x = a in the real number line (i.e. the x axis) but
it never crosses x = a , then it is written as $\mathrm{x}\to {\mathrm{a}}^{+}$.

Similarly,

3. NEIGHBOURHOOD OF ‘x = a ’
For some h > 0, sufficiently small, let the function y=f(x) be defined in the interval (a − h, a) then it is said that the function y = f(x) is defined in the left-neighbourhood of x = a.

Similarly, if the function y = f(x) be defined in the interval (a, a + h) then it is said that the function
y = f(x) is defined in the right-neighbourhood of x = a.

If the function y = f(x) be defined in left-neighbourhood of x = a or right-neighbourhood of x = a,
then it is said that the function y = f(x) is defined in the neighbourhood of x = a .

It must be noted here that the value ‘a’ itself may or may not be included in the domain which is
actually not being considered in its neighbourhood.

4. INDETERMINATE FORMS
Some times we come across some functions which do not have definite value corresponding to some particular value of the variable.

For example, the function f(x) = $\frac{{\mathrm{x}}^2–4}{\mathrm{x}–2}$, converts into $\frac{0}{0}$ if x = 2 is substituted.

Hence, f (2) cannot be determined. Such a form is called an Indeterminate Form.

There are total 7 Indeterminate Forms given as

(1) $\frac{0}{0}$,   (2) $\frac{\infty }{\infty }$,   (3) $\infty , –\infty$, (4) $0\times \infty$,   (5) $1^{\infty }$,   (6) ${\infty }^0$,   (7) $0^0$

Note : Here 0 and 1 are all approaching values, not the exact values.

Illustration-1
Which of the following are forming indeterminate form. Also indicate the form

(i) $\frac{1}{\mathrm{x}} \mathrm{as} \mathrm{x}\to 0$

(ii) $\frac{1–\mathrm{x}}{1–{\mathrm{x}}^2}\text{ as }\mathrm{x}\to 1$

(iii) $\mathrm{x}\ell \mathrm{n}\text{ xas }\mathrm{x}\to 0$

(iv) $\left(\frac{1}{\mathrm{x}}–\frac{1}{{\mathrm{x}}^2}\right)\text{ as }\mathrm{x}\to 0$

(v) $(\mathrm{sinx}{)}^{\mathrm{x}}\text{ as }\mathrm{x}\to 0$

(vi) $(\mathcal{l}\mathrm{nx}{)}^{\mathrm{x}}\text{ as }\mathrm{x}\to 0$

(vii) $(1+\mathrm{sinx}{)}^{\frac{1}{\mathrm{x}}}\text{ as }\mathrm{x}\to 0$

(viii) $\frac{\mathrm{secx}}{\mathrm{tanx}}\text{ as }\mathrm{x}\to \frac{\mathrm{\pi }}{2}$

Solution

(i) No

(ii) $\frac{0}{0}$ form

(iii) 0 × $\infty$ form

(iv) $\left(\infty , –\infty \right)$ form

(v) (0)º form

(vi) ($\infty$)º form

(vii) ${\left(1\right)}^{\infty }$ form

(viii) $\frac{\infty }{\infty }$ form

5. LIMIT OF A FUNCTION
Definition-1
Let the function y = f(x) be defined in a certain neighbourhood of a point x = a . The function y = f(x) approaches the limit L (y $\to$ L) as x approaches ‘a’ (x $\to$ a). If for every positive number h, arbitrarily small, we are able to indicate k > 0, arbitrarily small, such that for all x, different from ‘a’ and satisfying the inequality.

$\left|\mathrm{x}–\mathrm{a}\right|<\mathrm{h}$

we have the inequality

$\left|\mathrm{f}\right(\mathrm{x})–\mathrm{L}|<\mathrm{k}$

then $\underset{\mathrm{x}\to 2}{\lim } \mathrm{f}\left(\mathrm{x}\right)=\mathrm{L}$

or f(x) $\to$ L as x $\to$ a or limiting value of f(x) is L as x $\to$ a.

Definition-2
Let y = f (x) be a function of x and the limiting value of y is required for x $\to$ a, then we consider the values of the function at the points which are very near to ‘a’.
If these values tend to a definite unique number L as x tends to ‘a’ (either from left or from right)
then this unique number L is called the limit of f(x) at x = a and we write it as $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{L}$

Illustration-2

Evaluate $\underset{\mathrm{x}\to 2}{\lim }(\mathrm{x}+2)$

Solution

x + 2 being a polynomial in x, its limit as x $\to$ 2 is given by $\underset{\mathrm{x}\to 2}{\lim }(\mathrm{x}+2)=2+2=4$

Illustration-3

Evaluate $\underset{\mathrm{x}\to 2}{\lim }\mathrm{x}(\mathrm{x}–1)$

Solution

x(x – 1) being a polynomial in x, its limit as x $\to$ 2 is given by $\underset{\mathrm{x}\to 2}{\lim }\mathrm{x}(\mathrm{x}–1)=2(2–1)=2$

Illustration-4

Evaluate $\underset{\mathrm{x}\to 2}{\lim }\frac{{\mathrm{x}}^2+4}{\mathrm{x}+2}$

Solution

$\underset{\mathrm{x}\to 2}{\lim }\frac{{\mathrm{x}}^2+4}{\mathrm{x}+2}=\frac{(2{)}^2+4}{2+2}=2$

Illustration-5

Evaluate $\underset{\mathrm{x}\to 0}{\lim } \cos \left(\mathrm{sinx}\right)$

Solution

$\underset{\mathrm{x}\to 0}{\lim }\cos \left(\mathrm{sinx}\right)=\cos \left(\underset{\mathrm{x}\to 0}{\lim }\mathrm{sinx}\right)=\cos 0=1$

Illustration-6

Find the limiting value of $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2–9}{\mathrm{x}–3}\text{ as }\mathrm{x}\to 3$.

Solution

At x = 3, $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2–9}{\mathrm{x}–3}$ converts into an indeterminate form of $\frac{0}{0}$.

Now when x tends to 3 from left or from right, it can be easily observed from the graph that the value of f(x) tends to 6.

Hence

$\underset{\mathrm{x}\to 3}{\lim }\frac{{\mathrm{x}}^2–9}{\mathrm{x}–3}=\underset{\mathrm{x}\to 3}{\lim }\frac{(\mathrm{x}–3)(\mathrm{x}+3)}{(\mathrm{x}–3)}$

$=\underset{\mathrm{x}\to 3}{\lim }(\mathrm{x}+3)=3+3=6$

6. LEFT HAND AND RIGHT HAND LIMIT

Consider the values of the functions at the points which are very near to a on the left of a. If these
values tend to a definite unique number as x tends to a, then the unique number so obtained is called left-hand limit of f(x) at x = a and symbolically we write it as $\mathrm{f}(\mathrm{a}–0)=\underset{\mathrm{x}\to {\mathrm{a}}^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\mathrm{f}(\mathrm{a}–\mathrm{h})$

which is expressed as $\mathrm{f}(\mathrm{a}+0)=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\mathrm{f}(\mathrm{a}+\mathrm{h})$.

7. EXISTENCE OF LIMIT

The limit of a function f(x) at a point x = a exists and equals to L if $\underset{\mathrm{x}\to {\mathrm{a}}^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}$ finite value, L.

Here $\underset{\mathrm{x}\to {\mathrm{a}}^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right) \mathrm{and} \underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ are called left hand limit (L.H.L.) and right hand limit (R.H.L.)
respectively.

Thus, if $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ exists then $\underset{\mathrm{x}\to {\mathrm{a}}^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}$ finite value, L $\Rightarrow$L.H.L. = R.H.L. = L

$\Rightarrow \underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{L}$

Illustration-7

The value of $\underset{\mathrm{x}\to 1}{\lim }\left[\mathrm{x}\right]$ is, where [ ] represents the greatest integer function.

(A) 1     (B) 2     (C) 4     (D) Does not exist

Solution

Left hand limit = $\underset{\mathrm{x}\to 1^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1^{–}}{\lim }\left[\mathrm{x}\right]=0$

and Right hand limit = $\underset{\mathrm{x}\to 1^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

= $\underset{\mathrm{x}\to 1^{+}}{\lim }\left[\mathrm{x}\right]=1$

$\because \underset{\mathrm{x}\to 1^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right)\ne \underset{\mathrm{x}\to 1^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

∴ limit does not exist.

Illustration-8

If $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}\frac{1}{1+{\mathrm{e}}^{–1/\mathrm{x}‘}}& \mathrm{x}\ne 0\\ 0,& \mathrm{x}=0\end{array}\right.$ then at x = 0

(A) right hand limit of f(x) exists but not left-hand limit
(B) left-hand limit of f(x) exists but not right- hand limit
(C) both limits exists but are not equal
(D) both limits exist and are equal

Solution

$\mathrm{f}\left(0^{–}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{1}{1+{\mathrm{e}}^{1/\mathrm{h}}}=\frac{1}{1+\infty }=0$

$\mathrm{f}\left(0^{+}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{1}{1+{\mathrm{e}}^{–1/\mathrm{h}}}=\frac{1}{1+0}=1$

∴ Both limits exist but are not equal.

Illustration-9

Find $\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

Solution
Left hand limit = 1
Right hand limit = 2
Hence $\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ does not exist.

Illustration-10

From the adjoint graph of y = f(x), find

(i) $\underset{\mathrm{x}\to 0}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

(ii) $\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

(iii) $\underset{\mathrm{x}\to 2}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

(iv) $\underset{\mathrm{x}\to 3}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

(v) $\underset{\mathrm{x}\to 4}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

Solution

(i) Here L.H.L. = 2 and R.H.L. = 1

$\therefore \underset{\mathrm{x}\to 0}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ does not exist

because left hand limit ≠ right hand limit.

(ii) $\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)=0$

(iii) $\underset{\mathrm{x}\to 2}{\lim }\mathrm{f}\left(\mathrm{x}\right)=1$

(iv) $\underset{\mathrm{x}\to 3}{\lim }\mathrm{f}\left(\mathrm{x}\right)=1$

(v) $\underset{\mathrm{x}\to 4}{\lim }\mathrm{f}\left(\mathrm{x}\right)=1$

8. EVALUATION OF LEFT HAND AND RIGHT HAND LIMITS

Right-hand limit means tendency of function when we approach x = a from the value which just greater than ‘a’ and we write $\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$.

Working rule to evaluate $\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

• Put x = a + h in f(x) to get $\underset{\mathrm{h}\to 0}{\lim }\mathrm{f}(\mathrm{a}+\mathrm{h})$

• Take the limit as h $\to$ 0

Left-hand limit means tendency of function when we approach x = a from the values which are just less than ‘a’ and we write $\underset{\mathrm{x}\to {\mathrm{a}}^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$.

Working rule to evaluate $\underset{\mathrm{x}\to {\mathrm{a}}^{–}}{\lim }\mathrm{f}\left(\mathrm{x}\right)$

• Put x = a – h in f(x) to get $\underset{\mathrm{h}\to 0}{\lim }\mathrm{f}(\mathrm{a}–\mathrm{h})$

• Take the limit as h $\to$ 0

Illustration-11

If $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}2\mathrm{x}+3\text{ if }\mathrm{x}\le 0\\ 3(\mathrm{x}+1)\text{ if }\mathrm{x}>0\end{array}\right.$.

Find left and right hand limits and choose whether f (x) has limit at the point x = 0.

Solution

$\begin{array}{r}\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}+3\text{ if }\mathrm{x}\le 0\\ 3(\mathrm{x}+1)\text{ if }\mathrm{x}>0\end{array}$

$\underset{x\to 0–}{Lt}–f\left(x\right)=\underset{x\to 0–}{Lt}(2x+3)=2\left(0\right)+3=3$

$\underset{\mathrm{x}\to 0+}{\mathrm{Lt}}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0+}{\mathrm{Lt}}3(\mathrm{x}+1)=3(0+1)=3$

$\underset{\mathrm{x}\to 0–}{Lt} \mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0+}{\mathrm{Lt}}\mathrm{f}\left(\mathrm{x}\right)=3$

$\underset{\mathrm{x}\to 0}{Lt}\mathrm{f}\left(\mathrm{x}\right)=3$

Illustration-12

If f(x) = $x^2$ if x $\le$1

                       x if 1 < x $\le$ 2

                       x–3 if x > 2

Find left and right hand limits and check whether f(x) has limit at the point x = 1 ; 2.

Solution

$\underset{\mathrm{x}\to 1–}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1–}{Lt}{\mathrm{x}}^{\mathrm{x}}=1$

$\underset{\mathrm{x}\to 1+}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1+}{Lt}{\mathrm{x}}^{\mathrm{x}}=1$

$\underset{\mathrm{x}\to 1–}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1+}{Lt}{\mathrm{x}}^{\mathrm{x}}=1$

$\underset{\mathrm{x}\to 1}{Lt}\mathrm{f}\left(\mathrm{x}\right)=1$

Now, $\underset{\mathrm{x}\to 2–}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 2–}{Lt}\mathrm{x}=2$

$\underset{\mathrm{x}\to 2+}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 2+}{Lt}\mathrm{x}–3=2–3=–1$

$\underset{\mathrm{x}\to 2–}{Lt}\mathrm{f}\left(\mathrm{x}\right)\ne \underset{\mathrm{x}\to 2+}{Lt}f\left(x\right)$

$\therefore \underset{\mathrm{x}\to 2}{Lt}\mathrm{f}\left(\mathrm{x}\right)$ does not exists.

3. DIFFERENCE BETWEEN THE VALUE OF A FUNCTION AT A POINT AND THE LIMIT AT A POINT

Case 1: $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left(\mathrm{x}\right)$ and f(a) both exist but are not equal.

Example

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\frac{{\mathrm{x}}^2–1}{\mathrm{x}–1},\begin{array}{c}\mathrm{x}\ne 1\\ \mathrm{x}=1\end{array}\right.$

$\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\lim }\frac{{\mathrm{x}}^2–1}{\mathrm{x}–1}=\underset{\mathrm{x}\to 1}{\lim }(\mathrm{x}+1)–2$

f(x) exists at x = 1

f(1) = 0, value of f also exist at x = 1

But $\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(1\right)$

Case 2: $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left(\mathrm{x}\right)$and f(a) both exist and are equal.

Example

f(x) = ${\mathrm{x}}^2$

$\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\lim }\left({\mathrm{x}}^2\right)=1$, limit exists, and f(1) = (1)2 = 1 $\Rightarrow$ Value of also exist.

$\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(1\right)$

4. THEOREMS ON LIMITS AND EVALUATION OF ALGEBRAIC LIMITS

1. FUNDAMENTAL THEOREMS ON LIMITS

The following theorems are very useful for evaluation of limits if $\underset{\mathrm{x}\to 0}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{l}\text{ and }\underset{\mathrm{x}\to 0}{\lim }\mathrm{g}\left(\mathrm{x}\right)=\mathrm{m}$ (l and m are real numbers) then

(1) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left(\mathrm{f}\right(\mathrm{x})+\mathrm{g}(\mathrm{x}\left)\right)=l+\mathrm{m}$         (Sum rule)

(2) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left(\mathrm{f}\right(\mathrm{x})–\mathrm{g}(\mathrm{x}\left)\right)=l–\mathrm{m}$        (Difference rule)

(3) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left(\mathrm{f}\right(\mathrm{x}).\mathrm{g}(\mathrm{x}\left)\right)=l.\mathrm{m}$              (Product rule)

(4) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{kf}\left(\mathrm{x}\right)=\mathrm{k}.l$                        (Constant multiple rule)

(5) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{l}{\mathrm{m}}, \mathrm{m}\ne 0$          (Quotient rule)

(6) If $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=+\infty \text{ or }–\infty ,\text{ then }\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}=0$

(7) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\log \left\{\mathrm{f}\right(\mathrm{x}\left)\right\}=\log \left\{\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)\right\}$

(8) If $\mathrm{f}\left(\mathrm{x}\right)\le \mathrm{g}\left(\mathrm{x}\right)\text{ for all }\mathrm{x},\text{ then }\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)\le \underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{g}\left(\mathrm{x}\right)$

(9) $\mathrm{f}\left(\mathrm{x}\right)\le \mathrm{g}\left(\mathrm{x}\right)\text{ for all }\mathrm{x},\text{ then }\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)\le \underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{g}\left(\mathrm{x}\right)$

(10) If p and q are integers, then $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left(\mathrm{f}\right(\mathrm{x}){)}^{\mathrm{p}/\mathrm{q}}=l^{\mathrm{p}/\mathrm{q}}$, provided $(l{)}^{p/q}$ is a real number.

(11) If $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{g}\right(\mathrm{x}\left)\right)=\mathrm{f}\left(\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{g}\left(\mathrm{x}\right)\right)=\mathrm{f}\left(\mathrm{m}\right)$ provided ‘f ’ is continuous at $\underset{x\to a}{\lim }\ln \left[f\right(x\left)\right]=\ln \left(l\right)$, only if l > 0.

2. EVALUATION OF LIMITS INVOLVING ALGEBRAIC FUNCTIONS.

To evaluate the limits involving algebraic functions we use the following methods:
1) Direct substitution method
2) Factorisation method
3) Rationalisation method
4) Application of the standard limit

$\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{{\mathrm{x}}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}–\mathrm{a}}=\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}$

1. Direct substitution method

This method can be used in the following cases :

i) If f(x) is a polynomial function, then $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{a}\right)$.

ii) If $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{P}\left(\mathrm{x}\right)}{\mathrm{Q}\left(\mathrm{x}\right)}$ where P(x) and Q(x) are polynomial functions then $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{P}\left(\mathrm{a}\right)}{\mathrm{Q}\left(\mathrm{a}\right)}$, provided Q(a) ≠ 0.

Illustration-13

Find $\underset{\mathrm{x}\to 1}{Lt}\left(2{\mathrm{x}}^2+3\mathrm{x}+4\right)$

Solution

$\underset{\mathrm{x}\to 1}{Lt}\left(2{\mathrm{x}}^2+3\mathrm{x}+4\right)=2(1{)}^2+3(1)+4=9$.

Illustration-14

Find $\underset{\mathrm{x}\to 1}{Lt}\frac{2\mathrm{x}–1}{3{\mathrm{x}}^2–4\mathrm{x}+5}$

Solution

$\underset{\mathrm{x}\to 1}{Lt}\frac{2\mathrm{x}–1}{3{\mathrm{x}}^2–4\mathrm{x}+5}=\frac{2\left(1\right)–1}{3(1{)}^2–4(1)+5}=\frac{1}{4}$.

2. Factiorisation Method

This method is used when $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}$ turns out to be an indeterminate form of the type $\frac{0}{0}$ by the substitution of x = a.

In such a case the numerator (Nr.) and the denominator (Dr.) are factorised and the common factor (x – a) is cancelled. After eliminating the common factor the substitution x = a gives the limit, if it exists.

Illustration-15

Evaluate $\underset{\mathrm{x}\to 2}{\text{Lt}}\frac{{\mathrm{x}}^2–5\mathrm{x}+6}{10–3\mathrm{x}–{\mathrm{x}}^2}$.

Solution

$\underset{\mathrm{x}\to 2}{\text{Lt}}\frac{{\mathrm{x}}^2–5\mathrm{x}+6}{10–3\mathrm{x}–{\mathrm{x}}^2}$

= $\underset{\mathrm{x}\to 2}{Lt}\frac{(\mathrm{x}–2)(\mathrm{x}–3)}{–(\mathrm{x}+5)(\mathrm{x}–2)}$

= $\underset{\mathrm{x}\to 2}{\mathrm{Lt}} \frac{\mathrm{x}–3}{–(\mathrm{x}+5)} (\because \mathrm{x}\ne 2)=\frac{1}{7}$

Illustration-16

Evaluate $\underset{\mathrm{x}\to 2}{Lt}\left(\frac{1}{\mathrm{x}–2}–\frac{4}{{\mathrm{x}}^2–4}\right)$.

Solution

$\underset{\mathrm{x}\to 2}{Lt}\left(\frac{1}{\mathrm{x}–2}–\frac{4}{{\mathrm{x}}^2–4}\right) (\infty –\infty \text{ form })$

$=Lt\left(\frac{\mathrm{x}–2}{{\mathrm{x}}^2–4}\right)\left(\frac{0}{0}\text{ form }\right)=\underset{\mathrm{x}\to 2}{Lt}\left(\frac{1}{\mathrm{x}+2}\right)=\frac{1}{4}$

(Cancelling the common factor (x–2)).

3. Rationalisation Method

This method is used when $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}$ is a $\frac{0}{0}$ form and either the Nr. or Dr. consists of expressions
involving radical signs.

Illustration-17

Show that $\underset{\mathrm{x}\to 1}{\mathrm{Lt}} \frac{\mathrm{x}–1}{\sqrt{{\mathrm{x}}^2+3}–2}=2$

Solution

$\underset{{}_{\mathrm{x}\to 1}}{Lt}\frac{\mathrm{x}–1}{\sqrt{{\mathrm{x}}^2+3}–2}=\underset{{}_{\mathrm{x}\to 1}}{Lt}\frac{(\mathrm{x}–1)\left[\sqrt{{\mathrm{x}}^2+3}+2\right]}{\left(\sqrt{{\mathrm{x}}^2+3}–2\right)\left(\sqrt{{\mathrm{x}}^2+3}+2\right)}$

$=\underset{\mathrm{x}\to 1}{Lt} \frac{(\mathrm{x}–1)\left[\sqrt{{\mathrm{x}}^2+3}+2\right)}{\left({\mathrm{x}}^2+3\right)–4}$ = $\underset{\mathrm{x}\to 1}{Lt} \frac{(\mathrm{x}–1)\left[\sqrt{{\mathrm{x}}^2+3}+2\right]}{{\mathrm{x}}^2–1}$ = $\underset{\mathrm{x}\to 1}{Lt} \frac{(\mathrm{x}–1)\left[\sqrt{{\mathrm{x}}^2+3}+2\right)}{(\mathrm{x}–1)(\mathrm{x}+1)} (\because \mathrm{x}\ne 1)$.

$=\underset{\mathrm{x}\to 1}{\mathrm{Lt}}\frac{\sqrt{{\mathrm{x}}^2+3}+2}{\mathrm{x}+1}=\frac{2+2}{2}=2$

Illustration-18

Evaluate $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sqrt{1+{\mathrm{x}}^3}–\sqrt{1–{\mathrm{x}}^3}}{{\mathrm{x}}^2}$

Solution

$\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sqrt{1+{\mathrm{x}}^3}–\sqrt{1–{\mathrm{x}}^3}}{{\mathrm{x}}^2}$ = $\underset{\mathrm{x}\to 0}{Lt}\frac{\left(\sqrt{1+{\mathrm{x}}^3}–\sqrt{1–{\mathrm{x}}^3}\right)\left(\sqrt{1+{\mathrm{x}}^3}+\sqrt{1–{\mathrm{x}}^3}\right)}{{\mathrm{x}}^2\left[\left(\sqrt{1+{\mathrm{x}}^3}+\sqrt{1–{\mathrm{x}}^3}\right)\right]}$

= $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\left(1+{\mathrm{x}}^3\right)–\left(1–{\mathrm{x}}^3\right)}{{\mathrm{x}}^2\left[\sqrt{1+{\mathrm{x}}^3}+\sqrt{1–{\mathrm{x}}^3}\right)}$

= $\underset{\mathrm{x}\to 0}{Lt}\frac{2{\mathrm{x}}^3}{{\mathrm{x}}^2\left[\sqrt{1+{\mathrm{x}}^3}+\sqrt{1–{\mathrm{x}}^3}\right]}$

= $\underset{\mathrm{x}\to 0}{Lt}\frac{2\mathrm{x}}{\left(\sqrt{1+{\mathrm{x}}^3}+\sqrt{1–{\mathrm{x}}^3}\right)}=\frac{2\left(0\right)}{\sqrt{1+0}+\sqrt{1–0}}=0$.

4) Application of the standard limit $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{{\mathrm{x}}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}–\mathrm{a}}=\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}$

This method is explained through the following examples.

Illustration-19

Compute $\underset{\mathrm{x}\to 3}{Lt}\frac{{\mathrm{x}}^4–81}{\mathrm{x}–3}$

Solution

$\underset{\mathrm{x}\to 3}{Lt}\frac{{\mathrm{x}}^4–81}{\mathrm{x}–3}=\underset{\mathrm{x}\to 3}{Lt}\frac{{\mathrm{x}}^4–3^4}{\mathrm{x}–3}(\mathrm{a}=3, \mathrm{n}=4)$

= $4.3^{4–1}=4\left(27\right)=108$.

Illustration-20

Find $\underset{\mathrm{x}\to 0}{Lt}\frac{(2+\mathrm{x}{)}^{\mathrm{n}}–2^{\mathrm{n}}}{\mathrm{x}}$

Solution

Let 2 + x = t. Then $\mathrm{x}\to 0\Rightarrow \mathrm{t}\to 2$.

$\underset{\mathrm{x}\to 0}{Lt}\frac{(2+\mathrm{x}{)}^{\mathrm{n}}–2^{\mathrm{n}}}{\mathrm{x}}=\underset{\mathrm{t}\to 2}{Lt}\frac{{\mathrm{t}}^{\mathrm{n}}–2^{\mathrm{n}}}{\mathrm{t}–2}=\mathrm{n}.2^{\mathrm{n}–1}$.

Illustration-21

Show that $\underset{\mathrm{x}\to 0}{Lt}\frac{(\mathrm{a}+\mathrm{x}{)}^{\mathrm{n}}–(\mathrm{a}–\mathrm{x}{)}^{\mathrm{n}}}{\mathrm{x}}=2.\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}$.

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{(\mathrm{a}+\mathrm{x}{)}^{\mathrm{n}}–(\mathrm{a}–\mathrm{x}{)}^{\mathrm{n}}}{\mathrm{x}}$

$=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left[\frac{(\mathrm{a}+\mathrm{x}{)}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}}+\frac{(\mathrm{a}–\mathrm{x}{)}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{–\mathrm{x}}\right]$

= $\underset{\left(\mathrm{i}\right)}{\underset{\mathrm{x}\to 0}{Lt}\frac{(\mathrm{a}+\mathrm{x}{)}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}}}+\underset{\left(\mathrm{ii}\right)}{\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{(\mathrm{a}–\mathrm{x}{)}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{–\mathrm{x}}}$

Put a + x = t in (i) and a – x = s in (ii) then x $\to$ 0 $\Rightarrow$ t $\to$ a, s $\to$ a

∴ Given limit = $\underset{\mathrm{t}\to \mathrm{a}}{Lt}\left(\frac{{\mathrm{t}}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{\mathrm{t}–\mathrm{a}}\right)+\underset{\mathrm{s}\to \mathrm{a}}{Lt}\left(\frac{{\mathrm{s}}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{\mathrm{s}–\mathrm{a}}\right)$

= $\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}+\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}=2\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}$.

Note:

i) $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{(\mathrm{a}+\mathrm{x}{)}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{\mathrm{x}}=\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}$.

ii) $\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{(\mathrm{a}–\mathrm{h}{)}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{–\mathrm{h}}=\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}$.

iii) $\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{(\mathrm{a}–\mathrm{h}{)}^{\mathrm{n}}–{\mathrm{a}}^{\mathrm{n}}}{\mathrm{h}}=–\mathrm{n}.{\mathrm{a}}^{\mathrm{n}–1}$.

Illustration-22

Evaluate $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{(1+\mathrm{x}{)}^{\frac{1}{8}}–(1–\mathrm{x}{)}^{\frac{1}{8}}}{\mathrm{x}}$.

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{(1+\mathrm{x}{)}^{\frac{1}{8}}–(1–\mathrm{x}{)}^{\frac{1}{8}}}{\mathrm{x}}$

= $\underset{\mathrm{x}\to 0}{Lt}\frac{(1+\mathrm{x}{)}^{\frac{1}{8}}–1}{\mathrm{x}}+\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{(1–\mathrm{x}{)}^{\frac{1}{8}}–1}{–\mathrm{x}}$

= $\frac{1}{8}(1{)}^{\frac{1}{8}–1}+\frac{1}{8}(1{)}^{\frac{1}{8}–1}=\frac{1}{4}$.

5. EVALUATION OF TRIGONOMETRIC LIMITS

To evaluate trigonometric limit the following results are very important.

(i) $\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{sinx}}{\mathrm{x}}=1=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}}{\mathrm{sinx}}$

(ii) $\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{tanx}}{\mathrm{x}}=1=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}}{\mathrm{tanx}}$

(iii) $\underset{\mathrm{x}\to 0}{\lim }\frac{{\sin }^{–1}\mathrm{x}}{\mathrm{x}}=1=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}}{{\sin }^{–1}\mathrm{x}}$

(iv) $\underset{\mathrm{x}\to 0}{\lim }\frac{{\tan }^{–1}\mathrm{x}}{\mathrm{x}}=1=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}}{{\tan }^{–1}\mathrm{x}}$

(v) $\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{sinx}}^0}{\mathrm{x}}=\frac{\mathrm{\pi }}{180}$

(vi) $\underset{\mathrm{x}\to 0}{\lim }\mathrm{cosx}=1$

(vii) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{\sin (\mathrm{x}–\mathrm{a})}{\mathrm{x}–\mathrm{a}}=1$

(viii) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{\tan (\mathrm{x}–\mathrm{a})}{\mathrm{x}–\mathrm{a}}=1$

(ix) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }{\sin }^{–1}\mathrm{x}={\sin }^{–1}\mathrm{a},\left|\mathrm{a}\right|\le 1$

(x) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }{\cos }^{–1}\mathrm{x}={\cos }^{–1}\mathrm{a}; \left|\mathrm{a}\right|\le 1$

(xi) $\underset{\mathrm{x}\to \mathrm{a}}{\lim }{\tan }^{–1}\mathrm{x}={\tan }^{–1}\mathrm{a};–\infty <\mathrm{a}<\infty$

(xii) $\underset{\mathrm{x}\to \infty }{\lim }\frac{\mathrm{sinx}}{\mathrm{x}}=\underset{\mathrm{x}\to \infty }{\lim }\frac{\mathrm{cosx}}{\mathrm{x}}=0$

(xiii) $\underset{\mathrm{x}\to \infty }{\lim }\frac{\sin (1/\mathrm{x})}{(1/\mathrm{x})}=1$

Illustration-23

Find $\underset{\mathrm{x}\to 0}{Lt}\frac{\sin 2\mathrm{x}}{\sin 3\mathrm{x}}$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{\sin 2\mathrm{x}}{\sin 3\mathrm{x}}=\underset{\mathrm{x}\to 0}{Lt}\left(\frac{\sin 2\mathrm{x}}{2\mathrm{x}}\right)\left(\frac{3\mathrm{x}}{\sin 3\mathrm{x}}\right)\frac{2}{3}$

= $1.1.\frac{2}{3}=\frac{2}{3}$

Note:

$\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{sinax}}{\mathrm{sinbx}}=\frac{\mathrm{a}}{\mathrm{b}}, \underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{sinax}–\mathrm{sinbx}}{\mathrm{x}}=\mathrm{a}–\mathrm{b}$.

Illustration-24

Find $\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{tanax}–\mathrm{tanbx}}{\mathrm{x}}$

Solution

$\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\mathrm{tanax}–\mathrm{tanbx}}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\mathrm{tanax}}{\mathrm{x}}–\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\mathrm{tanbx}}{\mathrm{x}}$

= $\underset{\mathrm{x}\to 0}{Lt}\frac{\tan \left(\mathrm{ax}\right)}{\mathrm{ax}}.\mathrm{a}–\underset{\mathrm{x}\to 0}{Lt}\frac{\tan \left(\mathrm{bx}\right)}{\mathrm{bx}}.\mathrm{b}$ = 1.a – 1. b = a – b.

Illustration-25

Evaluate

(i) $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sin \left({\mathrm{x}}^0\right)}{\mathrm{x}}$

(ii) $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sin \left({\mathrm{x}}^0\right)}{{\mathrm{x}}^0}$

Solution

(i) $1^0=\frac{\mathrm{\pi }}{180}$ radians $\Rightarrow {\mathrm{x}}^0=\frac{\mathrm{\pi } \mathrm{x}}{180}$ radians

$\therefore \underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sin \left({\mathrm{x}}^0\right)}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sin \left(\frac{\mathrm{\pi }}{180}\right)}{\mathrm{x}}$ = $\underset{\mathrm{x}\to 0}{Lt}\frac{\sin \left(\frac{\mathrm{\pi } \mathrm{x}}{180}\right)}{\left(\frac{\mathrm{\pi } \mathrm{x}}{180}\right)}.\frac{\mathrm{\pi }}{180}=1.\frac{\mathrm{\pi }}{180}=\frac{\mathrm{\pi }}{180}$

(ii) $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sin \left({\mathrm{x}}^0\right)}{{\mathrm{x}}^0}=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sin \left(\frac{\mathrm{\pi } \mathrm{x}}{180}\right)}{\left(\frac{\mathrm{\pi } \mathrm{x}}{180}\right)}=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\mathrm{sint}}{\mathrm{t}}=1$.

(Where t = $\frac{\mathrm{\pi x}}{100}$ and x $\to$ 0 $\Rightarrow$ t $\to$ 0)

Illustration-26

Find $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\sin (\mathrm{x}–\mathrm{a})+\tan (\mathrm{x}–\mathrm{a})}{{\mathrm{x}}^2–{\mathrm{a}}^2}$

Solution

Let x – a = t then x $\to$ a $\Rightarrow$ t $\to$ 0; x + a = t + 2a

$\therefore \underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\sin (\mathrm{x}–\mathrm{a})+\tan (\mathrm{x}–\mathrm{a})}{{\mathrm{x}}^2–{\mathrm{a}}^2}$

$=\underset{\mathrm{t}\to 0}{Lt}\frac{\mathrm{sint}+\mathrm{tant}}{\mathrm{t}(\mathrm{t}+2\mathrm{a})}=\underset{\mathrm{t}\to 0}{Lt}\left(\frac{\mathrm{sint}}{\mathrm{t}}+\frac{\mathrm{tant}}{\mathrm{t}}\right)\left(\frac{1}{\mathrm{t}+2\mathrm{a}}\right)$

= $(1+1)\frac{1}{2\mathrm{a}}=\frac{1}{\mathrm{a}}$.

Illustration-27

Evaluate $\underset{\mathrm{x}\to 0}{Lt}\frac{cosec\mathrm{x}–\mathrm{cotx}}{\mathrm{x}}$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{cosec\mathrm{x}–\mathrm{cotx}}{\mathrm{x}}$ $=\underset{\mathrm{x}\to 0}{Lt}\frac{1}{\mathrm{x}}\left[\frac{1–\mathrm{cosx}}{\mathrm{sinx}}\right]$

$\underset{\mathrm{x}\to 0}{Lim}\frac{1}{\mathrm{x}}\left(\frac{2{\sin }^2\mathrm{x}/2}{2\mathrm{sinx}/2 \mathrm{cosx}/2}\right)$

$=\underset{\mathrm{x}\to 0}{Lim}\frac{\mathrm{tanx}/2}{\mathrm{x}}=\frac{1}{2}$

Illustration-28

Evaluate $\underset{\mathrm{x}\to \mathrm{\pi }/2}{\mathrm{Lt}}\frac{1–{\sin }^3\mathrm{x}}{{\cos }^2\mathrm{x}}=$

Solution

$\underset{\mathrm{x}\to \mathrm{\pi }/2}{\mathrm{Lt}}\frac{1–{\sin }^3\mathrm{x}}{{\cos }^2\mathrm{x}}=\underset{\mathrm{x}\to \mathrm{\pi }/2}{\mathrm{Lt}}\frac{1+{\sin }^2\mathrm{x}+\mathrm{sinx}}{1+\mathrm{sinx}}$ $=\frac{1+1+1}{1+1}=\frac{3}{2}$

Illustration-29

Evaluate $\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{tanx}–\mathrm{sinx}}{{\mathrm{x}}^3}$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{tanx}–\mathrm{sinx}}{{\mathrm{x}}^3}=\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{sinx}\left[\frac{1}{\mathrm{cosx}}–1\right]}{{\mathrm{x}}^3}$.

= $\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{sinx}\left[\frac{1–\mathrm{cosx}}{\mathrm{cosx}}\right]}{{\mathrm{x}}^3}=\underset{\mathrm{x}\to 0}{Lt}\left(\frac{\mathrm{tanx}}{\mathrm{x}}\right)\left(\frac{2{\sin }^2\frac{\mathrm{x}}{2}}{{\mathrm{x}}^2}\right)$ = $1\left(2\times {\left(\frac{1}{2}\right)}^2\right)=\frac{1}{2}$.

Illustration-30

Compute $\underset{\mathrm{x}\to 0}{Lt}(cosec\mathrm{x}–\mathrm{cotx})$

Solution

$\underset{\mathrm{x}\to 0}{Lt}(cosec\mathrm{x}–\mathrm{cotx})$ = $\underset{\mathrm{x}\to 0}{Lt}\frac{1–\mathrm{cosx}}{\mathrm{sinx}}$

$=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{2{\sin }^2\left(\frac{\mathrm{x}}{2}\right)}{2.\sin \left(\frac{\mathrm{x}}{2}\right).\cos \left(\frac{\mathrm{x}}{2}\right)}$ = $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\tan \left(\frac{\mathrm{x}}{2}\right)=\tan \left(0\right)=0$.

Illustration-31

Find $\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\mathrm{Lt}} \frac{\mathrm{cosx}}{\left(\frac{\mathrm{\pi }}{2}–\mathrm{x}\right)}$

Solution

Put $\frac{\mathrm{\pi }}{2}–\mathrm{x}=\mathrm{t}$ then $\mathrm{x}\to \frac{\mathrm{\pi }}{2}\Rightarrow \mathrm{t}\to 0$

$\therefore \underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{Lt} \frac{\mathrm{cosx}}{\left({\displaystyle \frac{\mathrm{\pi }}{2}}–\mathrm{x}\right)}=\underset{\mathrm{t}\to 0}{Lt}\frac{\cos \left(\frac{\mathrm{\pi }}{2}–\mathrm{t}\right)}{\mathrm{t}}$

= $\underset{\mathrm{t}\to 0}{\mathrm{Lt}} \frac{\sin \mathrm{t}}{\mathrm{t}}=1$.

Illustration-32

Show that $\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{sinax}}{\mathrm{x}}=\mathrm{a}(\mathrm{a}\in \mathrm{R})$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{sinax}}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{Lt}\frac{\sin \left(\mathrm{ax}\right)}{\mathrm{ax}}.\mathrm{a}$

= $\mathrm{a}.\underset{\mathrm{t}\to 0}{\mathrm{Lt}} \left(\frac{\mathrm{sint}}{\mathrm{t}}\right)=\mathrm{a}.1$ = a (where t = ax)

Illustration-33

Show that $\underset{\mathrm{x}\to 0}{Lt}\frac{{\sin }^{–1}\left(\mathrm{x}\right)}{\mathrm{x}}=1$

Solution

Let ${\sin }^{–1}\left(\mathrm{x}\right)=\mathrm{t}$ so that x = sint, then x $\to$ 0 $\Rightarrow$ t $\to$ 0

$\therefore \underset{\mathrm{x}\to 0}{\mathrm{Lt}} \frac{{\sin }^{–1}\left(\mathrm{x}\right)}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{\mathrm{Lt}} \frac{\mathrm{t}}{\mathrm{sint}}=1$.

Illustration-34

Show that $\underset{\mathrm{x}\to 0}{Lt}\frac{1–\mathrm{cosax}}{1–\mathrm{cosbx}}=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{1–\mathrm{cosax}}{1–\mathrm{cosbx}}=\underset{\mathrm{x}\to 0}{Lt}\frac{2{\sin }^2\left(\frac{\mathrm{ax}}{2}\right)}{2{\sin }^2\left(\frac{\mathrm{bx}}{2}\right)}$ = $\underset{\mathrm{x}\to 0}{Lt}{\left[\frac{{\displaystyle \raisebox{1ex}{$\sin \left(\frac{\mathrm{ax}}{2}\right)$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}}{{\displaystyle \raisebox{1ex}{$\sin \left({\displaystyle \frac{\mathrm{bx}}{2}}\right)$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.}}\right]}^2=\frac{{\left(\frac{\mathrm{a}}{2}\right)}^2}{{\left(\frac{\mathrm{b}}{2}\right)}^2}=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}$.

Illustration-35

Show that $\underset{\mathrm{x}\to 0}{\mathrm{Lt}} \frac{\left|\mathrm{sinx}\right|}{\mathrm{x}}$ does not exist.

Solution

$\left|\mathrm{sinx}\right|=\left\{\begin{array}{lll}–\mathrm{sinx}& \text{ if }& –\frac{\mathrm{\pi }}{2}<\mathrm{x}<0\\ +\mathrm{sinx}& \text{ if }& 0<\mathrm{x}<\mathrm{\pi }\end{array}\right.$

$\therefore \underset{\mathrm{x}\to 0–}{\mathrm{Lt}}\frac{\left|\mathrm{sinx}\right|}{\mathrm{x}}={Lt}_{\mathrm{x}\to 0–}\frac{–\mathrm{sinx}}{\mathrm{x}}=–1$

$\therefore \underset{\mathrm{x}\to 0+}{\mathrm{Lt}}\frac{\left|\mathrm{sinx}\right|}{\mathrm{x}}=\underset{\mathrm{x}\to 0+}{\mathrm{Lt}}\frac{\mathrm{sinx}}{\mathrm{x}}=1$

As x $\to$ 0, LHL ≠ RHL

$\Rightarrow \underset{\mathrm{x}\to 0}{Lt}\frac{\left|\mathrm{sinx}\right|}{\mathrm{x}}$ does not exist.

6. EVALUATION OF EXPONENTIAL AND LOGARITHMIC LIMITS

1. LOGARITHMIC LIMITS

To evaluate the logarithmic limits we use following formulae

(i) $\log (1+\mathrm{x})=\mathrm{x}–\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}–\dots \dots \dots \dots .\text{ to ∞}$ where $–1<x\le 1$ and expansion is true only if base is e.

(ii) $\underset{\mathrm{x}\to 0}{\lim }\frac{\log (1+\mathrm{x})}{\mathrm{x}}=1$

(iii) $\underset{\mathrm{x}\to \mathrm{e}}{\lim }{\log }_{\mathrm{e}}\mathrm{x}=1$

(iv) $\underset{\mathrm{x}\to 0}{\lim }\frac{\log (1–\mathrm{x})}{\mathrm{x}}=–1$

(v) $\underset{\mathrm{x}\to 0}{\lim }\frac{{\log }_{\mathrm{a}}(1+\mathrm{x})}{\mathrm{x}}={\log }_{\mathrm{a}}\mathrm{e}, \mathrm{a}>0, \ne 1$

2. EXPONENTIAL LIMITS
i. Based on series expansion

We use ${\mathrm{e}}^{\mathrm{x}}=1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+\dots \dots \dots \dots .\infty$

To evaluate the exponential limits we use the following results

(a) $\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}–1}{\mathrm{x}}=1$

(b) $\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{a}}^{\mathrm{x}}–1}{\mathrm{x}}={\log }_{\mathrm{e}}\mathrm{a}$

(c) $\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{\lambda x}}–1}{\mathrm{x}}=\mathrm{\lambda }(\mathrm{\lambda }\ne 0)$

ii. Based on the form $1^{\infty }$ :

To evaluate the exponential form $1^{\infty }$ we use the following results.

(a) If $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{g}\left(\mathrm{x}\right)=0$, then

$\underset{\mathrm{x}\to \mathrm{a}}{\lim }\{1+\mathrm{f}(\mathrm{x}){\}}^{1/\mathrm{g}\left(\mathrm{x}\right)}={\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}}$ or when $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=1$ and $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\mathrm{g}\left(\mathrm{x}\right)=\infty$.

Then $\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left\{\mathrm{f}\right(\mathrm{x}){\}}^{\mathrm{g}\left(\mathrm{x}\right)}=\underset{\mathrm{x}\to \mathrm{a}}{\lim }[1+\mathrm{f}\left(\mathrm{x}\right)–1{]}^{\mathrm{g}\left(\mathrm{x}\right)}$ = ${\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{a}}{\lim }\left(\mathrm{f}\right(\mathrm{x})–1)\mathrm{g}\left(\mathrm{x}\right)}$

(b) $\underset{\mathrm{x}\to 0}{\lim }(1+\mathrm{x}{)}^{1/\mathrm{x}}=\mathrm{e}$

(c) $\underset{\mathrm{x}\to \infty }{\lim }{\left(1+\frac{1}{\mathrm{x}}\right)}^{\mathrm{x}}=\mathrm{e}$

(d) $\underset{\mathrm{x}\to 0}{\lim }(1+\mathrm{\lambda x}{)}^{1/\mathrm{x}}={\mathrm{e}}^{\mathrm{\lambda }}$

(e) $\underset{\mathrm{x}\to \infty }{\lim }{\left(1+\frac{\mathrm{\lambda }}{\mathrm{x}}\right)}^{\mathrm{x}}={\mathrm{e}}^{\mathrm{\lambda }}$

• $\underset{\mathrm{x}\to \infty }{\lim }{\mathrm{a}}^{\mathrm{x}}=\left\{\begin{array}{l}\infty ,\text{ if }\mathrm{a}>1\\ 0,\text{ if }\mathrm{a}<1\end{array}\text{ i.e., }{\mathrm{a}}^{\infty }=\infty \right.$, if a > 1 and ${\mathrm{a}}^{\infty }=0$ if a < 1.

Illustration-36

Evaluate $\underset{\mathrm{x}\to 0}{Lt}\frac{{27}^{\mathrm{x}}–9^{\mathrm{x}}–3^{\mathrm{x}}+1}{{\mathrm{x}}^2}$.

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{{27}^{\mathrm{x}}–9^{\mathrm{x}}–3^{\mathrm{x}}+1}{{\mathrm{x}}^2}$ = $\underset{\mathrm{x}\to 0}{Lt}\frac{\left(9^{\mathrm{x}}–1\right)\left(3^{\mathrm{x}}–1\right)}{{\mathrm{x}}^2}$

= $\underset{\mathrm{x}\to 0}{Lt}\left(\frac{9^{\mathrm{x}}–1}{\mathrm{x}}\right).\underset{\mathrm{x}\to 0}{Lt}\left(\frac{3^{\mathrm{x}}–1}{\mathrm{x}}\right)$ = log 9. log3

Illustration-37

Evaluate $\underset{\mathrm{x}\to 0}{Lt}\left(\frac{{\mathrm{e}}^{{\mathrm{x}}^2}–\mathrm{cosx}}{{\mathrm{x}}^2}\right)$.

Solution

$\underset{\mathrm{x}\to 0}{Lt}\left(\frac{{\mathrm{e}}^{{\mathrm{x}}^2}–\mathrm{cosx}}{{\mathrm{x}}^2}\right)=\underset{\mathrm{x}\to 0}{Lt}\frac{\left({\mathrm{e}}^{\mathrm{x}}–1\right)+(1–\mathrm{cosx})}{{\mathrm{x}}^2}$

= $\underset{\mathrm{x}\to 0}{Lt}\left(\frac{{\mathrm{e}}^{{\mathrm{x}}^2}–1}{{\mathrm{x}}^2}\right)+\underset{\mathrm{x}\to 0}{Lt}\left(\frac{1–\mathrm{cosx}}{{\mathrm{x}}^2}\right)$ = $1+\frac{1}{2}=\frac{3}{2}$.

Illustration-38

Evaluate $\underset{\mathrm{x}\to 0}{Lt}(1+\mathrm{ax}{)}^{\mathrm{b}/\mathrm{x}}$ where a, b are constants.

Solution

We know that $\underset{\mathrm{t}\to 0}{\mathrm{Lt}}(1+\mathrm{t}{)}^{\mathrm{t}}=\mathrm{e}$.

Now, $\underset{\mathrm{x}\to 0}{Lt}(1+\mathrm{ax}{)}^{\mathrm{b}/\mathrm{x}}=\underset{\mathrm{x}\to 0}{Lt}{\left[(1+\mathrm{ax}{)}^{\frac{1}{\mathrm{ax}}}\right]}^{\mathrm{ax}·\frac{\mathrm{b}}{\mathrm{x}}}$

${\left[\underset{\mathrm{x}\to 0}{\mathrm{Lt}}(1+\mathrm{ax}{)}^{\frac{1}{\mathrm{ax}}}\right]}^{\mathrm{ab}}={\left[\underset{\mathrm{t}\to 0}{\mathrm{Lt}}(1+\mathrm{t}{)}^{\frac{1}{\mathrm{t}}}\right]}^{\mathrm{ab}}$, (where t = ax)

= ${\mathrm{e}}^{\mathrm{ab}}$

Aliter:

The given limit is of the form $1^{\infty }$.

$\therefore \underset{\mathrm{x}\to 0}{Lt}(1+\mathrm{ax})\frac{\mathrm{b}}{\mathrm{x}}={\mathrm{e}}^{\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\mathrm{b}}{\mathrm{x}}\left[1+\mathrm{ax}–1\right]}={\mathrm{e}}^{\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left(\mathrm{ax}\right)}={\mathrm{e}}^{\mathrm{ab}}.$

Illustration-39

Evaluate $\underset{\mathrm{x}\to \infty }{Lt}{\left(1+\frac{1}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{cx}+\mathrm{d}}$ where a,b,c,d are positive constants.

Solution

$\underset{\mathrm{x}\to \infty }{Lt}{\left(1+\frac{1}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{cx}+\mathrm{d}}$ ($1^{\infty }$ form)

$=\underset{\mathrm{x}\to \infty }{\mathrm{Lt}}{\left[{\left(1+\frac{1}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{ax}+\mathrm{b}}\right]}^{\frac{\mathrm{cx}+\mathrm{d}}{\mathrm{ax}+\mathrm{b}}}$ = ${\left[\underset{\mathrm{x}\to \infty }{Lt}{\left(1+\frac{1}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{ax}+\mathrm{b}}\right]}^{\underset{\mathrm{x}\to \infty }{\mathrm{Lt}}\frac{\mathrm{cx}+\mathrm{d}}{\mathrm{ax}+\mathrm{b}}}$ = ${\mathrm{e}}^{\raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{b}$}\right.}$

Aliter: $\underset{\mathrm{x}\to \infty }{Lt}{\left(1+\frac{1}{\mathrm{ax}+\mathrm{b}}\right)}^{\mathrm{cx}+\mathrm{d}}$ = ${\mathrm{e}}^{\underset{\mathrm{x}\to \infty }{Lt}(\mathrm{cx}+\mathrm{d})}\left[1+\frac{1}{\mathrm{ax}+\mathrm{b}}–1\right]$

= ${\mathrm{e}}^{\underset{\mathrm{x}\to \infty }{\mathrm{Lt}}\frac{\mathrm{cx}+\mathrm{d}}{\mathrm{ax}+\mathrm{b}}}={\mathrm{e}}^{\mathrm{c}/\mathrm{a}}$

Illustration-40

Find $\underset{\mathrm{x}\to 0}{Lt}{\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}^{\frac{\mathrm{sinx}}{\mathrm{x}–\mathrm{sinx}}}$

Solution

The given limit is of $1^{\infty }$ form

$\therefore \underset{\mathrm{x}\to 0}{Lt}{\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}^{\frac{\mathrm{sinx}}{\mathrm{x}–\mathrm{sinx}}}$ = ${\mathrm{e}}^{\underset{\mathrm{x}\to 0}{Lt}\left(\frac{\mathrm{sinx}}{\mathrm{x}–\mathrm{sinx}}\right)\left[\frac{\mathrm{sinx}}{\mathrm{x}}–1\right]}$ = ${\mathrm{e}}^{\underset{\mathrm{x}\to 0}{Lt}\left(–\frac{\mathrm{sinx}}{\mathrm{x}}\right)}={\mathrm{e}}^{–1}$.

Illustration-41

Find $\underset{\mathrm{x}\to \infty }{Lt}{\left(\frac{\mathrm{x}+\mathrm{a}}{\mathrm{x}+\mathrm{b}}\right)}^{\mathrm{x}}$, a, b are constants.

Solution

$\underset{\mathrm{x}\to \infty }{Lt}{\left(\frac{\mathrm{x}+\mathrm{a}}{\mathrm{x}+\mathrm{b}}\right)}^{\mathrm{x}}$ = $\underset{\mathrm{x}\to \infty }{Lt}{\left[1+\left(\frac{\mathrm{x}+\mathrm{a}}{\mathrm{x}–\mathrm{b}}–1\right)\right]}^{\mathrm{x}}$

= $\underset{\mathrm{x}\to \infty }{Lt}{\left[1+\frac{\mathrm{a}–\mathrm{b}}{\mathrm{x}+\mathrm{b}}\right]}^{\mathrm{x}} = \underset{\mathrm{x}\to \infty }{Lt}{\left[{\left(1+\frac{\mathrm{a}–\mathrm{b}}{\mathrm{x}+\mathrm{b}}\right)}^{\frac{\mathrm{x}+\mathrm{b}}{\mathrm{a}–\mathrm{b}}}\right]}^{\frac{\left(\mathrm{a}–\mathrm{b}\right)\mathrm{x}}{\mathrm{x}+\mathrm{b}}}$

= $\underset{\mathrm{x}\to \infty }{Lt}{\left[{\left(1+\frac{1}{\mathrm{t}}\right)}^{\mathrm{t}}\right]}^{\underset{\mathrm{x}\to \infty }{Lt}\frac{\left(\mathrm{a}–\mathrm{b}\right)\mathrm{x}}{\mathrm{x}+\mathrm{b}}}$, where $\mathrm{t}=\frac{\mathrm{x}+\mathrm{b}}{\mathrm{a}–\mathrm{b}}$

= ${\mathrm{e}}^{\mathrm{a}–\mathrm{b}}$.

Illustration-42

Evaluate $\underset{\mathrm{x}\to \infty }{Lt}{\left(\frac{{\mathrm{x}}^2+\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2+\mathrm{cx}+\mathrm{d}}\right)}^{\mathrm{x}}$ where a,b,c,d are constants.

Solution

$\underset{\mathrm{x}\to \infty }{Lt}{\left(\frac{{\mathrm{x}}^2+\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2+\mathrm{cx}+\mathrm{d}}\right)}^{\mathrm{x}} \left(1^{\infty } – \mathrm{form}\right)$

= ${\mathrm{e}}^{\underset{\mathrm{x}\to \infty }{Lt}\left[\frac{{\mathrm{x}}^2+\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2+\mathrm{cx}+\mathrm{d}}–1\right]}={\mathrm{e}}^{\underset{\mathrm{x}\to \infty }{Lt}\left[\frac{\mathrm{x}\left[\right(\mathrm{a}–\mathrm{c})\mathrm{x}+(\mathrm{b}–\mathrm{d}\left)\right]}{{\mathrm{x}}^2+\mathrm{cx}+\mathrm{d}}\right]}$

= ${\mathrm{e}}^{\underset{\mathrm{x}\to \infty }{\mathrm{Lt}}\left[\frac{\left[(\mathrm{a}–\mathrm{c})+\frac{(\mathrm{b}–\mathrm{d})}{\mathrm{x}}\right]}{1+\frac{\mathrm{c}}{\mathrm{x}}+\frac{\mathrm{d}}{{\mathrm{x}}^2}}\right]}={\mathrm{e}}^{\mathrm{a}–\mathrm{c}}$

Illustration-43

Evaluate $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}{\left(\frac{{\mathrm{a}}^{\mathrm{x}}+{\mathrm{b}}^{\mathrm{x}}+{\mathrm{c}}^{\mathrm{x}}}{3}\right)}^{\frac{1}{\mathrm{x}}}$ where a,b,c $\in {\mathrm{R}}^{+}$

Solution

The given limit is of $1^{\infty }–\mathrm{form}$

$\underset{\mathrm{x}\to \infty }{\mathrm{Lt}}{\left(\frac{{\mathrm{a}}^{\mathrm{x}}+{\mathrm{b}}^{\mathrm{x}}+{\mathrm{c}}^{\mathrm{x}}}{3}\right)}^{\frac{1}{\mathrm{x}}}$

= ${\mathrm{e}}^{\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{1}{\mathrm{x}}\left[\frac{{\mathrm{a}}^{\mathrm{x}}+{\mathrm{b}}^{\mathrm{x}}+{\mathrm{c}}^{\mathrm{x}}}{3}–1\right]}={\mathrm{e}}^{\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{1}{3}\left[\frac{{\mathrm{a}}^{\mathrm{x}}+{\mathrm{b}}^{\mathrm{x}}+{\mathrm{c}}^{\mathrm{x}}–3}{3}\right]}$

= ${\mathrm{e}}^{\frac{1}{3}.\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left[\frac{{\mathrm{a}}^{\mathrm{x}}–1}{\mathrm{x}}+\frac{{\mathrm{b}}^{\mathrm{x}}–1}{\mathrm{x}}+\frac{{\mathrm{c}}^{\mathrm{x}}–1}{\mathrm{x}}\right]}$

= ${\mathrm{e}}^{\frac{1}{3}\left(\mathrm{loga}+\mathrm{logb}+\mathrm{logc}\right)}={\mathrm{e}}^{\mathcal{l}\mathrm{n}\left(\sqrt[3]{\mathrm{abc}}\right)}=\sqrt[3]{\mathrm{abc}}$.

Note :

i) $\underset{\mathrm{x}\to 0}{Lt}{\left(\frac{1^{\mathrm{x}}+2^{\mathrm{x}}+\dots +{\mathrm{n}}^{\mathrm{x}}}{\mathrm{n}}\right)}^{\frac{1}{\mathrm{x}}}=\sqrt[\mathrm{n}]{\mathrm{n}!}$.

ii) $\underset{\mathrm{x}\to \infty }{Lt}{\left(\frac{{\mathrm{a}}^{\frac{1}{\mathrm{x}}}+{\mathrm{b}}^{\frac{1}{\mathrm{x}}}+{\mathrm{c}}^{\frac{1}{\mathrm{x}}}}{3}\right)}^{\mathrm{x}}=\sqrt[3]{\mathrm{abc}}$.

Illustration-44

If $\underset{\mathrm{x}\to 0}{Lt}{\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)}^{2/\mathrm{x}}={\mathrm{e}}^3$, find a and b.

Solution

${\mathrm{e}}^3=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}{\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.} \left(1^{\infty }–\mathrm{form}\right)$

= ${\mathrm{e}}^{\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{2}{\mathrm{x}}\left[1+\mathrm{ax}+{\mathrm{bx}}^2–1\right]}$

= ${\mathrm{e}}^{2.\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left[\frac{\mathrm{ax}+{\mathrm{bx}}^2}{\mathrm{x}}\right]} = {\mathrm{e}}^{2.\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left[\mathrm{a}+\mathrm{bx}\right]}$ = ${\mathrm{e}}^{2\mathrm{a}}$, for any $\mathrm{b}\in \mathrm{R}$

$\therefore {\mathrm{e}}^3={\mathrm{e}}^{2\mathrm{a}}\text{ and }\mathrm{b}\in \mathrm{R}$ $\Rightarrow \mathrm{a}=\frac{3}{2}, \mathrm{b}\in \mathrm{R}$

7. EVALUATION OF LIMITS BY USING DE’L’ HOSPITAL’S RULE

1) Let f(x) and g(x) be two functions such that $\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{Lt}}\mathrm{f}\left(\mathrm{x}\right)=0\text{ and }\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{Lt}}\mathrm{g}\left(\mathrm{x}\right)=0$

Then $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{Lt}}\frac{{\mathrm{f}}^{‘}\left(\mathrm{x}\right)}{{\mathrm{g}}^{‘}\left(\mathrm{x}\right)}$ provided the latter limit exists. (Here ‘ denotes differentiation wrt x)

2) Let f(x) and g(x) be two functions such that $\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{Lt}}\mathrm{f}\left(\mathrm{x}\right)=\infty \text{ and }\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{Lt}}\mathrm{g}\left(\mathrm{x}\right)=\infty$.

Then $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{Lt}}\frac{{\mathrm{f}}^{‘}\left(\mathrm{x}\right)}{{\mathrm{g}}^{‘}\left(\mathrm{x}\right)}$ provided the latter limit exists.

Note:

i) L’ Hospital’s rule can be applied repeatedly i.e., $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\mathrm{f}\left(\mathrm{x}\right)}{g\left(\mathrm{x}\right)}=\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{{\mathrm{f}}^{‘}\left(\mathrm{x}\right)}{{\mathrm{g}}^{‘}\left(\mathrm{x}\right)}=\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{{\mathrm{f}}^{‘‘}\left(\mathrm{x}\right)}{{\mathrm{g}}^{‘‘}\left(\mathrm{x}\right)}=...$ provided these limits exist and at each stage of application of the rule, we should make sure that the limit is either a $\frac{0}{0}$ from or an $\frac{\infty }{\infty }$ form.

ii) While applying the L’Hospital’ s rule the derivatives of the Numerator f(x) and the Denominator g(x) w.r.t x are to be calculated separately and at the same time, but not by using quotient rule on f(x) / g(x).

iii) L’Hospital’s rule is directly applicable to $\frac{1}{3}$ and $\frac{\infty }{\infty }$ forms only. This rule is not applicable directly to other indeterminate forms. $0.\infty ,\infty –\infty ,1^{\infty },0°\text{ and }\infty °$ but can be applied only after transforming them into either $\frac{0}{0}$ form or $\frac{\infty }{\infty }$ form.

Illustration-45

Find $\underset{\mathrm{x}\to 0}{Lt}\frac{1–\mathrm{cosx}}{{\mathrm{x}}^2}$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{1–\mathrm{cosx}}{{\mathrm{x}}^2}\left(\frac{0}{0}–\text{ form }\right)=\underset{\mathrm{x}\to 0}{Lt}\left(\frac{0+\mathrm{sinx}}{2\mathrm{x}}\right)$

$=\frac{1}{2}\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{sinx}}{\mathrm{x}}\left(\frac{0}{0}–\text{ form }\right)$

$=\frac{1}{2}\underset{\mathrm{x}\to 0}{Lt}\left(\frac{\mathrm{cosx}}{1}\right)=\frac{1}{2}\cos \left(0\right)=\frac{1}{2}$.

Illustration-46

Evaluate $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{{\mathrm{x}}^{\mathrm{a}}–{\mathrm{a}}^{\mathrm{x}}}{{\mathrm{x}}^{\mathrm{x}}–{\mathrm{a}}^{\mathrm{a}}}$.

Solution

$\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{{\mathrm{x}}^{\mathrm{a}}–{\mathrm{a}}^{\mathrm{x}}}{{\mathrm{x}}^{\mathrm{x}}–{\mathrm{a}}^{\mathrm{a}}}\left(\frac{0}{0}–\text{ form }\right)=\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\mathrm{a}.{\mathrm{x}}^{\mathrm{a}–1}–{\mathrm{a}}^{\mathrm{x}}\mathrm{loga}}{{\mathrm{x}}^{\mathrm{x}}(1+\mathrm{logx})–0}$

$=\frac{\mathrm{a}.{\mathrm{a}}^{\mathrm{a}–1}–{\mathrm{a}}^{\mathrm{a}}\mathrm{loga}}{{\mathrm{a}}^{\mathrm{a}}(1+\mathrm{loga})}=\frac{1–\mathrm{loga}}{1+\mathrm{loga}}=\frac{\log \left(\frac{\mathrm{e}}{\mathrm{a}}\right)}{\log \left(\mathrm{ea}\right)}$.

Illustration-47

Find $\underset{\mathrm{x}\to \infty }{Lt}\frac{\mathrm{logx}}{\mathrm{x}}$.

Solution

$\underset{\mathrm{x}\to \infty }{Lt}\frac{\mathrm{logx}}{\mathrm{x}}\left(\frac{\infty }{\infty }\text{ form }\right)·=\underset{\mathrm{x}\to \infty }{Lt}\frac{\left(\frac{1}{\mathrm{x}}\right)}{1}=\underset{\mathrm{x}\to \infty }{Lt}\left(\frac{1}{\mathrm{x}}\right)=0$.

Illustration-48

If $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\cos 4\mathrm{x}+\mathrm{acos}2\mathrm{x}+\mathrm{b}}{{\mathrm{x}}^4}$ is finite find the values of a and b. Also find the limit.

Solution

Let $\mathcal{l}=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\cos 4\mathrm{x}+\mathrm{acos}2\mathrm{x}+\mathrm{b}}{{\mathrm{x}}^4}=\frac{1+\mathrm{a}+\mathrm{b}}{0}$

This can be finite only if 1 + a + b = 0 — (1)

$\mathrm{\ell }=\underset{\mathrm{x}\to 0}{Lt}\frac{–4\sin 4\mathrm{x}–2\mathrm{asinx}}{4{\mathrm{x}}^3}$ (applying L’ Hospital’s rule)

$=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{–16\cos 4\mathrm{x}–4\mathrm{acos}2\mathrm{x}}{12{\mathrm{x}}^2}$ (applying L’ Hospital’s rule) = $\frac{–16–4a}{0}$

This can be finite only if – 16 – 4a = 0 — (2)
From (2), a = – 4
From (1), b = 3 $\therefore$ a = –4, b = 3.

Now, required limit is $\mathrm{\ell }=\underset{\mathrm{x}\to 0}{Lt}\frac{–16\cos 4\mathrm{x}+16\cos 2\mathrm{x}}{12{\mathrm{x}}^2}$

$=\underset{\mathrm{x}\to 0}{Lt}\left(\frac{–4}{3}\right)\left(\frac{\cos 4\mathrm{x}+\cos 2\mathrm{x}}{{\mathrm{x}}^2}\right)\left(\frac{0}{0}–\text{ form }\right)$ = $–\frac{4}{3}\underset{\mathrm{x}\to 0}{Lt}\frac{–4\sin 4\mathrm{x}–2\sin 2\mathrm{x}}{2\mathrm{x}}$

$=\frac{4}{3}\underset{\mathrm{x}\to 0}{Lt}\left[2\left(\frac{\sin 4\mathrm{x}}{\mathrm{x}}\right)+\left(\frac{\sin 2\mathrm{x}}{\mathrm{x}}\right)\right]=\frac{4}{3}(8+2)=\frac{40}{3}$.

8. SOME USEFUL RESULTS AND FREQUENTLY USED EXPANSIONS

1. SOME USEFUL RESULTS

i. Let S = {x,sin x, tan x,sinh x, tanh x,sin^-1 x, tan^-1 x,sinh^-1 x, tanh^-1 x}

If f (x), g(x)$\in$S then $\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{f}\left(\mathrm{mx}\right)}{\mathrm{g}\left(\mathrm{nx}\right)}=\frac{\mathrm{m}}{\mathrm{n}}$

If ${\mathrm{f}}_1\left(\mathrm{x}\right), {\mathrm{f}}_2\left(\mathrm{x}\right), {\mathrm{g}}_1\left(\mathrm{x}\right), {\mathrm{g}}_2\left(\mathrm{x}\right)\in \mathrm{S}$ then $\underset{\mathrm{x}\to 0}{Lt}\frac{{\mathrm{f}}_1\left(\mathrm{mx}\right)\pm {\mathrm{f}}_2\left(\mathrm{nx}\right)}{{\mathrm{g}}_1\left(\mathrm{px}\right)\pm {\mathrm{g}}_2\left(\mathrm{qx}\right)}=\frac{\mathrm{m}\pm \mathrm{n}}{\mathrm{p}\pm \mathrm{q}}$

ii. $\underset{\mathrm{x}\to 0}{Lt}\frac{\sin \mathrm{ax}}{\tan \mathrm{bx}}=\frac{\mathrm{a}}{\mathrm{b}},\underset{\mathrm{x}\to 0}{Lt}\frac{1–\cos \mathrm{ax}}{{\mathrm{x}}^2}=\frac{{\mathrm{a}}^2}{2}$

2. EVALUATION OF LIMITS USING SERIES EXPANSIONS

In the evaluation of certain limits it becomes necessary to use the following series expansions:

i) ${\mathrm{e}}^{\mathrm{x}}=1+\frac{\mathrm{x}}{1!}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+\dots .(\mathrm{x}\in \mathrm{R})$

ii) ${\mathrm{a}}^{\mathrm{x}}=1+\frac{\mathrm{x}}{1!}\left(\mathcal{l}\mathrm{na}\right)+\frac{{\mathrm{x}}^2}{2!}(\mathrm{\ell na}{)}^2+\frac{{\mathrm{x}}^3}{3!}(\mathcal{l}\mathrm{na}{)}^3$+….(a>0, x$\in$R)

iii) $\mathrm{\ell n}(1+\mathrm{x})=\mathrm{x}–\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}–\frac{{\mathrm{x}}^4}{4}+\dots ,\left|\mathrm{x}\right|<1$

iv) $(1+\mathrm{x}{)}^{\mathrm{n}}=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}–1)}{2!}{\mathrm{x}}^2+\frac{\mathrm{n}(\mathrm{n}–1)(\mathrm{n}–2)}{3!}{\mathrm{x}}^3+\dots .(\mathrm{n}\in \mathrm{Q},\left|\mathrm{x}\right|<1)$

v) $\mathrm{sinx}=\mathrm{x}–\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}–\frac{{\mathrm{x}}^7}{7!}+\dots . (\mathrm{x}\in \mathrm{R})$

vi) $\mathrm{cosx}=1–\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}–\frac{{\mathrm{x}}^6}{6!}+\dots . (\mathrm{x}\in \mathrm{R})$

vii) $\mathrm{tanx}=\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\frac{2}{15}{\mathrm{x}}^5+\frac{17}{315}{\mathrm{x}}^7+\dots$

viii) $\mathrm{sinhx}=\mathrm{x}+\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}+\dots . (\mathrm{x}\in \mathrm{R})$

ix) $\mathrm{coshx}=1+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}+\frac{{\mathrm{x}}^6}{6!}+\dots (\mathrm{x}\in \mathrm{R})$

x) $(1+\mathrm{x}{)}^{1/\mathrm{x}}=\mathrm{e}\left[1–\frac{\mathrm{x}}{2}+\frac{11}{24}{\mathrm{x}}^2+\dots \right]$

xi) ${\sin }^{–1}\mathrm{x}=\mathrm{x}+1^2.\frac{{\mathrm{x}}^3}{3!}+\frac{1^2.3^2.{\mathrm{x}}^5}{5!}+1^2.3^2.5^2.\frac{{\mathrm{x}}^7}{7!}+\dots$

xii) ${\tan }^{–1}\mathrm{x}=\mathrm{x}–\frac{1}{3}{\mathrm{x}}^3+\frac{1}{5}{\mathrm{x}}^5–\dots$

xiii) ${\sec }^{–1}\mathrm{x}=1+\frac{{\mathrm{x}}^2}{2!}+5·\frac{{\mathrm{x}}^4}{4!}+61\frac{{\mathrm{x}}^6}{6!}+\dots$

Illustration-49

Find $\underset{\mathrm{x}\to 0}{Lt}\left[\frac{\mathrm{sinx}–\mathrm{x}+\frac{{\mathrm{x}}^3}{6}}{{\mathrm{x}}^5}\right]$.

Solution

$\underset{\mathrm{x}\to 0}{Lt}\left[\frac{\mathrm{sinx}–\mathrm{x}+\frac{{\mathrm{x}}^3}{6}}{{\mathrm{x}}^5}\right]$

$=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\left(\mathrm{x}–\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}–\frac{{\mathrm{x}}^7}{7!}+\dots \right)–\mathrm{x}+\frac{{\mathrm{x}}^3}{6}}{{\mathrm{x}}^5}$

$=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\left(\frac{{\mathrm{x}}^5}{5!}–\frac{{\mathrm{x}}^7}{7!}+\dots \right)}{{\mathrm{x}}^5}$

$=\underset{\mathrm{x}\to 0}{Lt}$ ($\frac{1}{5!}–\frac{{\mathrm{x}}^2}{7!}+$terms containing positive integral powers of x)

$=\frac{1}{5!}=\frac{1}{120}$.

Illustration-50

Find the values of a,b and c if $\underset{\mathrm{x}\to 0}{Lt}\left(\frac{{\mathrm{ae}}^{\mathrm{x}}–\mathrm{bcosx}+{\mathrm{ce}}^{–\mathrm{x}}}{{\mathrm{x}}^2}\right)=2$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\left(\frac{{\mathrm{ae}}^{\mathrm{x}}–\mathrm{bcosx}+{\mathrm{ce}}^{–\mathrm{x}}}{{\mathrm{x}}^2}\right)=$$\underset{\mathrm{x}\to 0}{Lt}\frac{\left[\begin{array}{l}a\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)\\ –b\left(1–\frac{x^2}{2!}+\frac{x^4}{4!}–+..\right)\\ +c\left(1–x+\frac{x^2}{2!}–\frac{x^3}{3!}+..\right)\end{array}\right]}{x^2}$

$=\underset{\mathrm{x}\to 0}{Lt}\left(\frac{1}{{\mathrm{x}}^2}\right)\left((\mathrm{a}–\mathrm{b}+\mathrm{c})+(\mathrm{a}–\mathrm{c})\mathrm{x}+\left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\right){\mathrm{x}}^2+(\dots .){\mathrm{x}}^3+...\right)$

$=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left[\frac{\mathrm{a}–\mathrm{b}+\mathrm{c}}{{\mathrm{x}}^2}+\frac{\mathrm{a}–\mathrm{c}}{\mathrm{x}}+\frac{(\mathrm{a}+\mathrm{b}+\mathrm{c})}{2}+\text{ terms }\mathrm{containing} \mathrm{positive} \mathrm{powers} \mathrm{of} \mathrm{x}\right]$

The limit is finite and is equal to 2 only if

a – b + c = 0 — (1),      a – c = 0 — (2)

and $\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=2$…..(3)

Solving (1) (2) and (3), a = 1, b = 2, c = 1.

Illustration-51

Evaluate $\underset{\mathrm{x}\to 0}{Lt}\left[\frac{(1+\mathrm{x}{)}^{\frac{1}{\mathrm{x}}}–\mathrm{e}+\frac{\mathrm{ex}}{2}}{{\sin }^2\mathrm{x}}\right]$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\frac{(1+\mathrm{x}{)}^{\frac{1}{\mathrm{x}}}–\mathrm{e}+\frac{\mathrm{ex}}{2}}{{\sin }^2\mathrm{x}}$ = $\underset{\mathrm{x}\to 0}{Lt}\frac{(1+\mathrm{x}{)}^{1/\mathrm{x}}–\mathrm{e}+\frac{\mathrm{ex}}{2}}{{\mathrm{x}}^2}.\left(\frac{{\mathrm{x}}^2}{{\sin }^2\mathrm{x}}\right)$

= $\underset{\mathrm{x}\to 0}{Lt}\frac{(1+\mathrm{x}{)}^{1/\mathrm{x}}–\mathrm{e}+\frac{\mathrm{ex}}{2}}{{\mathrm{x}}^2}.1$ = $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\mathrm{e}\left[1–\frac{\mathrm{x}}{2}+\frac{11}{24}{\mathrm{x}}^2+\dots \right]–\mathrm{e}+\frac{\mathrm{ex}}{2}}{{\mathrm{x}}^2}$

= $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}$ ($\frac{11\mathrm{e}}{24}$+term containing positive powers of x)

= $\frac{11\mathrm{e}}{24}$.

9. SANDWICH THEOREM OR SQUEEZE PRINCIPLE

If $\mathrm{f}\left(\mathrm{x}\right)\le \mathrm{g}\left(\mathrm{x}\right)\le \mathrm{h}\left(\mathrm{x}\right)\forall \mathrm{x}$ & $\underset{\mathrm{x}\to \mathrm{a}}{\text{Limit}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{\ell }=\underset{\mathrm{x}\to \mathrm{a}}{Limit}\mathrm{h}\left(\mathrm{x}\right)$ then $\underset{\mathrm{x}\to \mathrm{a}}{\text{Limit}}\mathrm{g}\left(\mathrm{x}\right)=\mathrm{\ell }$

Illustration-52

Evaluate $\underset{\mathrm{n}\to \infty }{\lim }\frac{\left[\mathrm{x}\right]+\left[2\mathrm{x}\right]+\left[3\mathrm{x}\right]+\dots .+\left[\mathrm{nx}\right]}{{\mathrm{n}}^2}$, Where [ ] denotes the greatest integer function.

Solution

We know that, x – 1 < |x| $\le$ x

$$\begin{gathered} \Rightarrow 2\mathrm{x}–1<\left[2\mathrm{x}\right]\le 2\mathrm{x} \\ \Rightarrow 3\mathrm{x}–1<\left[3\mathrm{x}\right]\le 3\mathrm{x} \\ ............................. \\ ............................. \end{gathered}$$

$\Rightarrow \mathrm{nx}–1<\left[\mathrm{nx}\right]\le \mathrm{nx}$

$\therefore$ (x + 2x + 3x + …. + nx) – n < [x] + [2x] + ….. +[nx] $\le$ (x + 2x + …. + nx)

$\Rightarrow \frac{\mathrm{xn}(\mathrm{n}+1)}{2}–\mathrm{n}<\sum _{\mathrm{r}=1}^{\mathrm{n}}\left[\mathrm{rx}\right]\le \frac{\mathrm{x}.\mathrm{n}(\mathrm{n}+1)}{2}$

Thus, $\underset{\mathrm{n}\to \infty }{\lim }\frac{\left[\mathrm{x}\right]+\left[2\mathrm{x}\right]+\dots .+\left[\mathrm{nx}\right]}{{\mathrm{n}}^2}$

$\Rightarrow \underset{\mathrm{n}\to \infty }{\lim }\frac{\mathrm{x}}{2}\left(1+\frac{1}{\mathrm{n}}\right)–\frac{1}{\mathrm{n}}$< $\underset{\mathrm{n}\to \infty }{\lim }\frac{\left[\mathrm{x}\right]+\left[2\mathrm{x}\right]+\dots .+\left[\mathrm{nx}\right]}{{\mathrm{n}}^2}\le \underset{\mathrm{n}\to \infty }{\lim }\frac{\mathrm{x}}{2}\left(1+\frac{1}{\mathrm{n}}\right)$

$\Rightarrow \frac{\mathrm{x}}{2}<\underset{\mathrm{n}\to \infty }{\lim }\frac{\left[\mathrm{x}\right]+\left[2\mathrm{x}\right]+\dots .+\left[\mathrm{nx}\right]}{{\mathrm{n}}^2}\le \frac{\mathrm{x}}{2}$$\Rightarrow \underset{\mathrm{n}\to \infty }{\lim }\frac{\left[\mathrm{x}\right]+\left[2\mathrm{x}\right]+\dots .+\left[\mathrm{nx}\right]}{{\mathrm{n}}^2}=\frac{\mathrm{x}}{2}$

Illustration-53

Find $\underset{\mathrm{x}\to \infty }{\mathrm{Lt}}\frac{[\mathrm{ax}+\mathrm{b}]}{\mathrm{x}}(\mathrm{a}>0)$.where [.] denotes the GIF.

Solution

For any x $\in$R, we know that x – 1 < [x] $\le$ x

$\therefore \mathrm{ax}+\mathrm{b}–1<[\mathrm{ax}+\mathrm{b}]\le \mathrm{ax}+\mathrm{b}$

$\Rightarrow \frac{\mathrm{ax}+\mathrm{b}–1}{\mathrm{x}}<\frac{[\mathrm{ax}+\mathrm{b}]}{\mathrm{x}}\le \frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}}$$\Rightarrow \underset{\mathrm{x}\to \infty }{\mathrm{Lt}}\frac{\mathrm{ax}+\mathrm{b}–1}{\mathrm{x}}\le \underset{\mathrm{x}\to \infty }{\mathrm{Lt}}\frac{[\mathrm{ax}+\mathrm{b}]}{\mathrm{x}}\le \underset{\mathrm{x}\to \infty }{\mathrm{Lt}}\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}}$

$\Rightarrow \mathrm{a}\le \underset{\mathrm{x}\to \infty }{Lt}\frac{[\mathrm{ax}+\mathrm{b}]}{\mathrm{x}}\le \mathrm{a}\Rightarrow \underset{\mathrm{x}\to \infty }{Lt}\frac{[\mathrm{ax}+\mathrm{b}]}{\mathrm{x}}=\mathrm{a}$.

10. MISCELLANEOUS CONCEPTS AND PROBLEMS

1. METHOD OF EVALUATING LIMITS OF ALGEBRAIC FUNCTIONS OF X WHEN $\mathit{X}\mathbf{\to }\mathbf{\pm }\mathbf{\infty }$

We make use of the following basic limits.

i) $\mathrm{x}\to \pm \infty \Rightarrow \frac{1}{\mathrm{x}}\to 0, \frac{\mathrm{k}}{\mathrm{x}}\to 0$ (k is a constant)

ii) If n>0 then $\underset{\mathrm{x}\to \infty }{\mathrm{Lt}}{\mathrm{x}}^{\mathrm{n}}=\infty , \underset{\mathrm{x}\to \infty }{\mathrm{Lt}}\frac{1}{{\mathrm{x}}^{\mathrm{n}}}=0$

iii) If $\mathrm{n}\in {\mathrm{Z}}^{+}$, then $\underset{\mathrm{x}\to –\infty }{\mathrm{Lt}}{\mathrm{x}}^{2\mathrm{n}}=+\infty ,\underset{\mathrm{x}\to –\infty }{\mathrm{Lt}}{\mathrm{x}}^{2\mathrm{n}+1}=–\infty$,

$\underset{\mathrm{x}\to –\infty }{Lt}\frac{1}{{\mathrm{x}}^{2\mathrm{n}}}=0, \underset{\mathrm{x}\to –\infty }{Lt}\frac{1}{{\mathrm{x}}^{2\mathrm{n}+1}}=0, \underset{\mathrm{x}\to –\infty }{Lt}\frac{1}{{\mathrm{x}}^{\mathrm{n}}}=0$

Illustration-54

Evaluate $\underset{\mathrm{x}\to \infty }{Lt}\frac{6{\mathrm{x}}^2–\mathrm{x}+7}{\mathrm{x}+3}$

Solution

$\underset{\mathrm{x}\to \infty }{Lt}\frac{6{\mathrm{x}}^2–\mathrm{x}+7}{\mathrm{x}+3}=\underset{\mathrm{x}\to \infty }{Lt}\frac{{\mathrm{x}}^2\left(6–\frac{1}{\mathrm{x}}+\frac{7}{{\mathrm{x}}^2}\right)}{\mathrm{x}\left(1+\frac{3}{\mathrm{x}}\right)}$

$=\underset{\mathrm{x}\to \infty }{Lt}\frac{\mathrm{x}\left(6–\frac{2}{\mathrm{x}}+\frac{7}{{\mathrm{x}}^2}\right)}{\left(1+\frac{3}{\mathrm{x}}\right)}=\frac{\infty .(6–0+0)}{(1+0)}=+\infty$

Illustration-55

Find $\underset{\mathrm{x}\to –\infty }{Lt}\frac{5{\mathrm{x}}^3+4}{\sqrt{2{\mathrm{x}}^4–1}}$

Solution

$\underset{\mathrm{x}\to –\infty }{Lt}\frac{5{\mathrm{x}}^3+4}{\sqrt{2{\mathrm{x}}^4+1}}=\underset{\mathrm{x}\to –\infty }{\mathrm{Lt}}\frac{{\mathrm{x}}^3\left(5+\frac{4}{{\mathrm{x}}^3}\right)}{{\mathrm{x}}^2\sqrt{2+\frac{1}{{\mathrm{x}}^4}}}$

$\underset{\mathrm{x}\to –\infty }{Lt}\frac{\mathrm{x}\left(5+\frac{4}{{\mathrm{x}}^3}\right)}{\sqrt{2+\frac{1}{{\mathrm{x}}^4}}}=(–\infty )·\frac{(5+0)}{\sqrt{2+0}}=–\infty$

2. EVALUATION OF INFINITE LIMITS

In evaluating this type of limits, we use the following basic results on limits :

i) $\underset{\mathrm{x}\to 0–}{Lt}\frac{1}{\mathrm{x}}=–\infty , \underset{\mathrm{x}\to 0+}{Lt}\frac{1}{\mathrm{x}}=+\infty , \underset{\mathrm{x}\to 0}{Lt}\frac{1}{\mathrm{x}}$ = does not exist.

ii) $\underset{\mathrm{x}\to \mathrm{a}–}{Lt}\frac{1}{\mathrm{x}–\mathrm{a}}=–\infty , \underset{\mathrm{x}\to \mathrm{a}+}{Lt}\frac{1}{\mathrm{x}–\mathrm{a}}=+\infty$, $\underset{\mathrm{x}\to \mathrm{a}–}{Lt}\frac{1}{\mathrm{a}–\mathrm{x}}=+\infty , \underset{\mathrm{x}\to \mathrm{a}+}{Lt}\frac{1}{\mathrm{a}–\mathrm{x}}=–\infty$

$\therefore {Lt}_{\mathrm{x}\to \mathrm{a}}\frac{1}{\mathrm{x}–\mathrm{a}}$ does not exist. For a positive integer, n

iii) $\underset{\mathrm{x}\to 0}{Lt}\frac{1}{{\mathrm{x}}^{2\mathrm{n}}}=+\infty , \underset{\mathrm{x}\to 0–}{Lt}\frac{1}{{\mathrm{x}}^{2\mathrm{n}+1}}=–\infty , \underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{1}{{\mathrm{x}}^{2\mathrm{n}+1}}=+\infty$

iv) $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{1}{(\mathrm{x}–\mathrm{a}{)}^{2\mathrm{n}}}=+\infty , \underset{\mathrm{x}\to \mathrm{a}–}{Lt}\frac{1}{(\mathrm{x}–\mathrm{a}{)}^{2\mathrm{n}+1}}=–\infty$, $\underset{\mathrm{x}\to \mathrm{a}+}{Lt}\frac{1}{(\mathrm{x}–\mathrm{a}{)}^{2\mathrm{n}+1}}=+\infty$.

v) $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{1}{(\mathrm{a}–\mathrm{x}{)}^{2\mathrm{n}}}=+\infty , \underset{\mathrm{x}\to \mathrm{a}–}{Lt}\frac{1}{(\mathrm{a}–\mathrm{x}{)}^{2\mathrm{n}+1}}=+\infty$, $\underset{\mathrm{x}\to \mathrm{a}+}{Lt}\frac{1}{(\mathrm{a}–\mathrm{x}{)}^{2\mathrm{n}+1}}=–\infty$.

vi) $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\infty \Rightarrow \underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}=0$.

vii) $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\mathrm{f}\left(\mathrm{x}\right)=–\infty \Rightarrow \underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}=0$.

viii) If $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\mathrm{f}\left(\mathrm{x}\right)=0$ and f(x) > 0 in a deleted nbd of a then $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}=+\infty$.

ix) If $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\mathrm{f}\left(\mathrm{x}\right)=0$ and f(x) < 0 in a deleted nbd of a then $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}=–\infty$.

x) If $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\mathrm{f}\left(\mathrm{x}\right)=0$ then $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}$ may or may not exist.

Illustration-56

Find $\underset{\mathrm{x}\to 3}{Lt}\frac{{\mathrm{x}}^2+3\mathrm{x}+2}{{\mathrm{x}}^2–6\mathrm{x}+9}$

Solution

Since $\underset{\mathrm{x}\to 3}{Lt}\left({\mathrm{x}}^2+3\mathrm{x}+2\right)=20$ and $\underset{\mathrm{x}\to 3}{Lt}\left({\mathrm{x}}^2–6\mathrm{x}+9\right)=0$

the given limit  $\underset{\mathrm{x}\to 3}{Lt}\frac{{\mathrm{x}}^2+3\mathrm{x}+2}{{\mathrm{x}}^2–6\mathrm{x}+9}$ can not be finite.

For x ≠ 3, ${\mathrm{x}}^2–6\mathrm{x}+9={\left(\mathrm{x}–3\right)}^2>0$ and ${\mathrm{x}}^2+3\mathrm{x}+2>0$ in a nbd, of the point 3.

Let $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2–6\mathrm{x}+9}{{\mathrm{x}}^2+3\mathrm{x}+2}$. Then f(x)>0 in a deleted nbd of 3 and $\underset{\mathrm{x}\to 3}{Lt}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 3}{Lt}\frac{(\mathrm{x}–3{)}^2}{{\mathrm{x}}^2+3\mathrm{x}+2}=0\Rightarrow \underset{\mathrm{x}\to 3}{Lt}\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}=+\infty$

Hence $\underset{\mathrm{x}\to 3}{Lt}\frac{{\mathrm{x}}^2+3\mathrm{x}+2}{{\mathrm{x}}^2–6\mathrm{x}+9}=+\infty$.

3. EVALUATION OF LIMITS OF FORM : $\infty –\infty , 0.\infty , 0^0 \mathrm{and} {\infty }^0$

i) If $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\left[\mathrm{f}\right(\mathrm{x})–\mathrm{g}(\mathrm{x}\left)\right]$ is of the form $\infty –\infty$, it can be transformed to $\frac{0}{0}–\mathrm{form}$ or $\frac{\infty }{\infty }$ form by writing it as $\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\left(\frac{1}{\mathrm{g}\left(\mathrm{x}\right)}\right)}\text{ or }\underset{\mathrm{x}\to \mathrm{a}}{Lt}\frac{\mathrm{g}\left(\mathrm{x}\right)}{\left(\frac{1}{\mathrm{f}\left(\mathrm{x}\right)}\right)}$ and hence can be evaluated.

iii) If $\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{Lt}} \mathrm{f}{\left(\mathrm{x}\right)}^{\mathrm{g}\left(\mathrm{x}\right)}$ is of the form $0^0$ or ${\infty }^0$ , it can be expressed as ${\mathrm{e}}^{\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{Lt}}\mathrm{g}\left(\mathrm{x}\right).\mathcal{ln}\mathrm{f}\left(\mathrm{x}\right)}$ so that the limit in the exponent is a 0. $\infty –\mathrm{form}$ and hence can be evaluated as in (ii).

Illustration-57

Determine $\underset{\mathrm{x}\to 0}{Lt}\left(\frac{1}{{\sin }^2\mathrm{x}}–\frac{1}{{\sinh }^2\mathrm{x}}\right)(\infty –\infty \text{ form })$

Solution

$\underset{\mathrm{x}\to 0}{Lt}\left(\frac{1}{{\sin }^2\mathrm{x}}–\frac{1}{{\sinh }^2\mathrm{x}}\right)=\underset{\mathrm{x}\to 0}{Lt}\left(\frac{{\sinh }^2\mathrm{x}–{\sin }^2\mathrm{x}}{{\sin }^2\mathrm{x}.{\sinh }^2\mathrm{x}}\right)$

$=\underset{\mathrm{x}\to 0}{Lt}\left[\frac{{\sinh }^2\mathrm{x}–{\sin }^2\mathrm{x}}{{\mathrm{x}}^4}.{\left(\frac{\mathrm{x}}{\mathrm{sinx}}\right)}^2.{\left(\frac{\mathrm{x}}{\mathrm{sinhx}}\right)}^2\right]$

$=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{{\sinh }^2\mathrm{x}–{\sin }^2\mathrm{x}}{{\mathrm{x}}^4}\left(\frac{0}{0}\text{ form }\right)$

$=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{\sinh 2\mathrm{x}–\sin 2\mathrm{x}}{4{\mathrm{x}}^3}\left(\frac{0}{0}\text{ form }\right)$

$=\underset{\mathrm{x}\to 0}{Lt}\frac{\cosh 2\mathrm{x}–\cos 2\mathrm{x}}{6{\mathrm{x}}^2}\left(\frac{0}{0}\text{ form }\right)$

$=\underset{\mathrm{x}\to 0}{Lt}\frac{1}{6}\left(\frac{\sinh 2\mathrm{x}+\sin 2\mathrm{x}}{\mathrm{x}}\right)=\frac{1}{6}(2+2)=\frac{2}{3}$.

Illustration-58

Find $\underset{\mathrm{x}\to 0}{Lt}\left(\mathrm{xlog}\right(\mathrm{tanx}\left)\right) (0.\infty –\text{ form })$

Solution

$\underset{\mathrm{x}\to 0}{Lt}(\mathrm{xlog} \mathrm{tanx})$= $\underset{\mathrm{x}\to 0}{Lt} \frac{\log \left(\mathrm{tanx}\right)}{\left({\displaystyle \frac{1}{\mathrm{x}}}\right)} \left(\frac{\infty }{\infty }\mathrm{form}\right)$

= $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left(\raisebox{1ex}{${\displaystyle \frac{{\sec }^2\mathrm{x}}{\mathrm{tanx}}}$}\!\left/ \!\raisebox{-1ex}{$\left({\displaystyle \frac{–1}{{\mathrm{x}}^2}}\right)$}\right.\right)=\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\frac{–{\mathrm{x}}^2{\sec }^2\mathrm{x}}{\mathrm{tanx}}$

= $\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left(–\mathrm{x}\right)\left(\frac{\mathrm{x}}{\mathrm{tanx}}\right).{\sec }^2\mathrm{x}=0.1.1=0$.

Illustration-59

Evaluate $\underset{\mathrm{x}\to 0}{Lt}\left({\mathrm{x}}^{\mathrm{x}}\right).\left(0^0\text{ form }\right)$.

Solution

$\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\left({\mathrm{x}}^{\mathrm{x}}\right)={\mathrm{e}}^{\underset{\mathrm{x}\to 0}{\mathrm{Lt}}\mathrm{x}.\mathrm{logx}}={\mathrm{e}}^{\mathcal{l}}$, say

Then $\mathrm{\ell }=\underset{\mathrm{x}\to 0}{Lt}\mathrm{xlogx}(0.\infty \text{ form })$

$=\underset{\mathrm{x}\to 0}{Lt}\frac{\mathrm{logx}}{\left(\frac{1}{\mathrm{x}}\right)} \left(\frac{\infty }{\infty }\text{ form }\right)$ = $\underset{\mathrm{x}\to 0}{Lt}\frac{\left(\frac{1}{\mathrm{x}}\right)}{\left(–\frac{1}{{\mathrm{x}}^2}\right)}=\underset{\mathrm{x}\to 0}{Lt}(–\mathrm{x})=0$

$\therefore \underset{\mathrm{x}\to 0}{\mathrm{Lt}}{\mathrm{x}}^{\mathrm{x}}={\mathrm{e}}^{\mathrm{\ell }}={\mathrm{e}}^0=1$.

Illustration-60

$\underset{\mathrm{x}\to 0+}{\mathrm{Lt}}{\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{tanx}} \left({\infty }^0 \mathrm{form}\right)$

Solution

Let $\mathrm{y}={\left(\frac{1}{\mathrm{x}}\right)}^{\mathrm{tanx}}$

Then log y = tan x log$\left(\frac{1}{\mathrm{x}}\right)\Rightarrow \mathrm{y}={\mathrm{e}}^{–\mathrm{tanx}.\mathrm{logx}}$

$\therefore \underset{\mathrm{x}\to 0+}{\mathrm{Lt}} \mathrm{y}={\mathrm{e}}^{–\underset{\mathrm{x}\to 0+}{\mathrm{Lt}}(\mathrm{tanx}·\mathrm{logx})}={\mathrm{e}}^{\mathrm{\ell }},\text{ say }$

Then, $\mathrm{\ell }=\underset{\mathrm{x}\to 0+}{\mathrm{Lt}}\mathrm{tanx}.\mathrm{logx}(0.\infty \text{ form })$

$=\underset{\mathrm{x}\to 0+}{Lt}\frac{\mathrm{logx}}{\mathrm{cotx}}\left(\frac{\infty }{\infty }\text{ form }\right)$ = $\underset{\mathrm{x}\to 0+}{Lt}\frac{\left({\displaystyle \frac{1}{\mathrm{x}}}\right)}{\left(–{cosec}^2\mathrm{x}\right)}=\underset{\mathrm{x}\to 0+}{Lt}(–\mathrm{sinx})\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)=0$

$\therefore \underset{\mathrm{x}\to 0+}{\mathrm{Lt}}\mathrm{y}={\mathrm{e}}^{–\mathrm{\ell }}={\mathrm{e}}^0=1$.

11. DERIVATIVES

We know that the tangent line to the curve y = f(x) at the point (a,f(a)) has slope

$\mathrm{m}=\mathrm{m}\left(\mathrm{a}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{a}+\mathrm{h})–\mathrm{f}\left(\mathrm{a}\right)}{\mathrm{h}}$

The derivative of the function f is the function f ‘ define by for all x for which this limit exist.

${\mathrm{f}}^{‘}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})–\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}$

Illustration-61

Find the derivative of $f\left(x\right)=\frac{1}{x}$.

Solution

We have ${\mathrm{f}}^{‘}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})–\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\frac{1}{(\mathrm{x}+\mathrm{h})}–\frac{1}{\mathrm{x}}}{\mathrm{h}}$ = $\underset{\mathrm{h}\to 0}{\lim }\frac{1}{\mathrm{h}}\left[\frac{\mathrm{x}–(\mathrm{x}+\mathrm{h})}{\mathrm{x}(\mathrm{x}+\mathrm{h})}\right]$

$=\underset{\mathrm{h}\to 0}{\lim }\frac{1}{\mathrm{h}}\left[\frac{–\mathrm{h}}{\mathrm{x}(\mathrm{x}+\mathrm{h})}\right]=\underset{\mathrm{h}\to 0}{\lim }\frac{–1}{\mathrm{x}(\mathrm{x}+\mathrm{h})}=–\frac{1}{{\mathrm{x}}^2}$

12. DERIVATIVES OF SOME OF THE FREQUENTLY USED FUNCTIONS

Function	Derivative
c (constant)	0
sinx	cosx
cos x	–sinx
tanx	sec2x
xn	nxn–1

The above written derivatives can be easily found by using the definition of differentiation.

13. RULES TO FIND OUT DERIVATIVES

Let u and v are differentiable functions of ‘x’.

(i) The sum rule

$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{u}+\mathrm{v})=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{dv}}{\mathrm{dx}}$

e.g. $\frac{\mathrm{d}}{\mathrm{dx}}\left(2{\mathrm{e}}^{\mathrm{x}}+3\mathrm{logx}\right)=2\frac{{\mathrm{de}}^{\mathrm{x}}}{\mathrm{dx}}+3\frac{\mathrm{d}\left(\mathrm{logx}\right)}{\mathrm{dx}}$$=2{\mathrm{e}}^{\mathrm{x}}+\frac{3}{\mathrm{x}}$

(ii) Product rule

$\frac{\mathrm{d}\left(\mathrm{uv}\right)}{\mathrm{dx}}=\mathrm{u}\frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v}\frac{\mathrm{du}}{\mathrm{dx}}$

e.g. $\frac{\mathrm{d}\left(\left(\mathrm{sinx}\right){\mathrm{e}}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{sinx}\frac{{\mathrm{de}}^{\mathrm{x}}}{\mathrm{dx}}+{\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}\left(\mathrm{sinx}\right)}{\mathrm{dx}}$

= (sinx) e^x + (cosx) e^x.

Illustration-62:

Differentiate ${\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}$.

Solution:

First we differentiate ${\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}$

$\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}\right)={\mathrm{x}}^2\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}\right)+{\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2\right)={\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}+2{\mathrm{xe}}^{\mathrm{x}}$

Now, $\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\right)={\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{sinx}\right)+\mathrm{sinx}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}\right)$

$={\mathrm{x}}^2{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}+\mathrm{sinx}\left({\mathrm{x}}^2+2\mathrm{x}\right){\mathrm{e}}^{\mathrm{x}}$

$={\mathrm{e}}^{\mathrm{x}}\left({\mathrm{x}}^2\mathrm{cosx}+{\mathrm{x}}^2\mathrm{sinx}+2\mathrm{xsinx}\right)={\mathrm{xe}}^{\mathrm{x}}(\mathrm{xcosx}+\mathrm{xsinx}+2\mathrm{sinx})$

(iii) The quotient rule

Here v(x) ≠ 0

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\frac{\mathrm{vdu}}{\mathrm{dx}}–\frac{\mathrm{udv}}{\mathrm{dx}}}{{\mathrm{v}}^2}$

e.g. $\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{tanx}}{\mathrm{x}}\right)=\frac{\mathrm{x}\frac{\mathrm{d}\left(\mathrm{tanx}\right)}{\mathrm{dx}}–\left(\mathrm{tanx}\right)\frac{\mathrm{dx}}{\mathrm{dx}}}{{\mathrm{x}}^2}$

$=\frac{{\mathrm{xsec}}^2\mathrm{x}–\mathrm{tanx}}{{\mathrm{x}}^2}$

Illustration-63:

Differentiate  $\frac{{\mathrm{e}}^{\mathrm{x}}}{1+\mathrm{sinx}}$

Solution:

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{e}}^{\mathrm{x}}}{1+\mathrm{sinx}}\right)=\frac{(1+\mathrm{sinx})\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}\right)–{\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}(1+\mathrm{sinx})}{(1+\mathrm{sinx}{)}^2}$

$=\frac{(1+\mathrm{sinx}){\mathrm{e}}^{\mathrm{x}}–{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}}{(1+\mathrm{sinx}{)}^2}=\frac{{\mathrm{e}}^{\mathrm{x}}(1+\mathrm{sinx}–\mathrm{cosx})}{(1+\mathrm{sinx}{)}^2}$

(iv) Chain rule

The chain rule is probably the most widely used differentiation rule in mathematics. Chain rule says that the derivative of the composite of two differentiable functions is the product of their derivatives evaluated at appropriate points.

The formula is $\left[\mathrm{f}\right(\mathrm{g}\left(\mathrm{x}\right))]‘=\mathrm{f}‘(\mathrm{g}\left(\mathrm{x}\right)).\mathrm{g}‘(\mathrm{x})$

Illustration-64:

Differentiate ${\mathrm{sinx}}^2$

Solution:

Put $\mathrm{y}={\mathrm{x}}^2$  and z = siny

Then $\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}\text{ and }\frac{\mathrm{dz}}{\mathrm{dy}}=\mathrm{cosy}$

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{sinx}}^2\right)=\frac{\mathrm{dz}}{\mathrm{dx}}=\frac{\mathrm{dz}}{\mathrm{dy}}.\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\mathrm{cosy}\right)\left(2\mathrm{x}\right)$ = $\left({\mathrm{cosx}}^2\right)\left(2\mathrm{x}\right)=2{\mathrm{xcosx}}^2$

This solution can be rewritten using a more convenient notation in the following manner:

$\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{sinx}}^2\right)=\frac{\mathrm{d}\left({\mathrm{sinx}}^2\right)}{\mathrm{d}\left({\mathrm{x}}^2\right)}.\frac{\mathrm{d}\left({\mathrm{x}}^2\right)}{\mathrm{dx}}$ = ${\mathrm{cosx}}^2.2\mathrm{x}=2{\mathrm{xcosx}}^2$

Illustration-65:

Differentiate $\sin \left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)+{\mathrm{e}}^{\sqrt{1+{\mathrm{x}}^2}}$  with respect to x.

Solution:

Let $\mathrm{y}=\sin \left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)+{\mathrm{e}}^{\sqrt{1+{\mathrm{x}}^2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\cos \left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\right)+{\mathrm{e}}^{\sqrt{1+{\mathrm{x}}^2}}\left(\frac{\mathrm{d}}{\mathrm{dx}}\sqrt{1+{\mathrm{x}}^2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\left(1+3{\mathrm{x}}^2\right)}{{\left(1+{\mathrm{x}}^2\right)}^2}\cos \left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)+\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}{\mathrm{e}}^{\sqrt{1+{\mathrm{x}}^2}}$