# Limits, Continuity and Differentiability - Evaluations and Examples

Table of Content

## What are Limits?

The expression $\underset{x\to c}{\mathop{\lim }}\,\,f(x)=L$ means that f(x) can be as close to L as desired by making x sufficiently close to ‘C’. In such a case, we say that the limit of f, as x approaches to C, is L.

Neighbourhood of a point

Let ‘a’ be real number and ‘h; is very close to ‘O’ then

Left hand limit will be obtained when x = a – h or x a

Similarly, Right Hand limit will be obtained when x = a + h or x a+

### Existence of limit

The limit will exist if the following conditions get fulfilled:

(a) $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,$ f(x) =$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,$ f(x) i.e. L.H.L = R.H.L

(b) Both L.H.L & R.H.L should be finite.

Examples

• $\underset{x\to 1}{\mathop{\lim }}\,\,{{x}^{2}}+1={{1}^{2}}+1=2$
• $\underset{x\to 0}{\mathop{\lim }}\,\,{{x}^{2}}-x={{0}^{2}}-0=0$
• $\underset{x\to 2}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4}{x+3}=\frac{4-4}{2+3}=0$
• In Limits, we have in determinant forms such as $\frac{0}{0},\frac{\infty }{\infty },0\times \infty ,\infty \times \infty ,{{1}^{\infty }},0{}^\circ ,\infty {}^\circ$

In these case, we try to simplify the problem into the valid function.

(c) Evaluate $\underset{x\to \infty }{\mathop{\lim }}\,\frac{x+7\sin x}{-2x+13}$using sandwich Theorem

Solution: $-1\le \sin x\le 1$ $\Rightarrow \,\,-7\le 7\sin x\le 7$ $\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\frac{x-7}{-2x+13}\ge \underset{x\to \infty }{\mathop{\lim }}\,\frac{x+7\sin x}{-2x+13}\ge \underset{x\to \infty }{\mathop{\lim }}\,\frac{x+7}{-2x+13}$

### Use of Expansion in Evaluating limits

Some Important Expansions

1. $\log (1+x)=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+…….$
2. ${{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}-\frac{{{x}^{4}}}{4!}+…….$
3. ${{a}^{x}}=1+x\log a+\frac{{{x}^{2}}}{2!}{(log\;a)}^{2}+…….$
4. $\sin x=x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}……$
5. $\cos x=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}……$
6. $\tan x=x+\frac{{{x}^{3}}}{3}+\frac{2}{15}{{x}^{5}}+……$

### Some Important limits

1. $\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{\sin x}{x}=1$
2. $\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{1-\cos x}{{{x}^{2}}}=\frac{1}{2}$
3. $\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{\tan x}{x}=1$
4. $\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{{{e}^{x}}-1}{x}=1$
5. $\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{\log (1+x)}{x}=1$

Example: $\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{\sin x-x}{{{x}^{3}}}$

Solution:

$\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\,\frac{\sin 2x}{x}$

### Evaluation of Algebric Limits

#### Direct substitution method

Example: $\underset{x\to 1}{\mathop{\lim }}\,\,(3{{x}^{2}}+4x+5)$= 3(1)2 + 4(1) + 5 = 12

Example: $\underset{x\to 2}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4}{x+3}=\frac{4-4}{2+3}=\frac{0}{5}=0$

#### Factorization method

Example: $\underset{x\to 2}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-5x+6}{{{x}^{2}}-4}$

Solution: $\underset{x\to 2}{\mathop{\lim }}\,\,\frac{(x-2)(x-3)}{(x+2)(x-2)}\Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\frac{x-3}{x+2}$= $\frac{-1}{4}$

#### Rationalization method

$\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sqrt{2+x}-\sqrt{2}}{x}$

Solution: $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\left( \sqrt{2+x}-\sqrt{2} \right)\left( \sqrt{2+x}+\sqrt{2} \right)}{x\left( \sqrt{2+x}+\sqrt{2} \right)}$

= $\frac{1}{2\sqrt{2}}$

Theorem: If nd then $\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$

Example: $\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{10}}-{{2}^{10}}}{{{x}^{5}}-{{2}^{5}}}$

Solution:

Limit of the form $0{}^\circ ,\infty {}^\circ$

Let L = $\underset{x\to a}{\mathop{lim}}\,{{\left( f(x) \right)}^{g(x)}}$

Example: $\underset{x-\infty }{\mathop{\lim }}\,\,\,\,{{x}^{{}^{1}/{}_{4}}}=?$

Solution: ${{e}^{\log }}\underset{x-\infty }{\mathop{\lim }}\,{{x}^{{}^{1}/{}_{x}}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,}}\,\,\,\frac{\log x}{x}$= e° = 1

$form\,{{1}^{\infty }}$ $\underset{x\to 0}{\mathop{\lim }}\,{{(1+x)}^{\frac{1}{x}}}=e\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,\,\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e$

Whenever we have ${{1}^{\infty }}$case, use ${{e}^{\underset{x\to a}{\mathop{\lim }}\,(f(x)-1)xg(x)}}$

Example: Evaluate $\underset{x\to 0}{\mathop{\lim }}\,{{(1+x)}^{cosecx}}$ $\underset{x\to 0}{\mathop{\lim }}\,{{\left[ {{(1+x)}^{\frac{1}{x}}} \right]}^{cosecx.x}}$= e1 = e

## Continuity

### What is Continuity

A continuous function is a function for which small changes in the input results in small changes in the output. Otherwise, a function is said to be discontinuous.

A function f(x) is said to be continuous at x = a if

$\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=f(a)$

i.e. L.H.L = R.H.L = value of the function at x = a

A function f(x) is said to be discontinuous Function.

Example 1: f(x) = $\frac{1}{2\sin x-1},$ Discuss the continuity or discontinuity

Solution: Clearly the function will be not defined for $\sin x=\frac{1}{2}=\sin \frac{\pi }{6}$

Function is discontinuous for $x=n\,\pi +{{(-1)}^{n}}\frac{\pi }{6}$

Example 2: What value must be assigned to K so that the function

$f(x)=\left\{\begin{matrix} \frac{x^4-256}{x-4} & , & x\neq 4\\ k & , & x=4 \end{matrix}\right.$ is continuous at x=4

Solution: =$f(4)=\underset{x\to 4}{\mathop{\lim }}\,\,\,\,\,\frac{{{x}^{4}}-256}{x-4}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{4}}-{{4}^{4}}}{x-4}={{4.4}^{4-1}}=256$

Example 3: Discuss the continuity of

(a) Sgn (x3x)

(b) f(x) = $\left[ \frac{2}{1+{{x}^{2}}} \right]$ , x > 0 [ ]

(a) f(x) = sgn (x3–x)

Here x3 – x = 0 x =0, –1, 1

Hence f(x) is discontinuous at x = 0, 1, –1

(b) $\frac{2}{1+{{x}^{2}}}$ , x > 0 is a monotonically decimal function

Hence f(x) = $\left[ \frac{2}{1+{{x}^{2}}} \right],\,\,\,x\ge 0$ is discontinuous,

When $\frac{2}{1+{{x}^{2}}}$ is on integer

$\Rightarrow \frac{2}{1+{{x}^{2}}}=1,\,2\,\,\,\,\,\,at\,\,\,\,x=1,0$

Example 4: Discuss the continuity of f(x) = $\left\{\begin{matrix} x-2 & x\leq 0\\ 4-x^2 &x>0 \end{matrix}\right.$ at x = 0

Solution: $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,\,x-2=-2$

### Intermediate Value Theorem

If f is continuous on [a, b] and f(a) ≠ f(b), then for any value $c\in (f(a),f(b))$, there is at least one number in x0 (a, b) for which f(x0) = c

Example: Show that the function f(x) = (xa)2 (xb)2 + x takes the values $\frac{a+b}{2}$ for some value of $x\in [a,b]$

Solution: f(a) = a, f(b) = b. Also f is continuous in [a, b] and $\frac{a+b}{2}$ $\in$ $\in$ [a, b]

Hence using I B I there exists at least one $c\in [a,b]$ such that f(c) = $\frac{a+b}{2}$

## Differentiability

### Existence of Derivative

Right and left hand derivative

F.H.D: f’(0+) =$\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,$ $\frac{f(a+h)-(a)}{h}$

L.H.D: F’$({{a}^{-}})$= $\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,$ $\frac{h(a-h)-f(a)}{-h}$

#### How can a function fail to be differentiable?

The function f(x) is said to be non-differentiable at x = a if

(a) Both R.H.D & L.H.D exist but not equal

(b) Either or both R.H.D & L.H.D are not finite

(c) Either or both R.H.D & L.H.D do not exist.

### Differentiability using first principle of Derivatives

$f(x)=\left\{\begin{matrix} \frac{sinx^3}{x} & x\neq 0\\ 0 & x=0 \end{matrix}\right.$ at x=0

f(x) = |x|, discuss the differentiability

Solution:

Clearly at x = 0, there is a sharp corner.

Hence the function is non-differentiable

f(x) = |x3|, Discuss the differentiability

f(x) = |x3| = $\left\{\begin{matrix} \ x^3 & ,x> 0\\ -x^3 & ,x<0 \end{matrix}\right.$

f’(x) = $\left\{\begin{matrix} \ 3x^2 & ,x> 0\\ -3x^2 & ,x<0 \end{matrix}\right.$

f’(0+) = f(0) = 0

Hence f(x) is differentiable at x = 0

#### Practise This Question

Find the area of triangle whose base is 4 cm long and the height corresponding to the base measures 12 cm.