 # Limits, Continuity and Differentiability - Evaluations and Examples

The limit concept is certainly indispensable for the development of analysis, for convergence and divergence of infinite series also depends on this concept.  The theory of limits and then defining continuity, differentiability and the definite integral in terms of the limit concept is successfully executed by mathematicians. In this section, will study this concept in detail with the help of solved examples.

## 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+

Related Concepts:

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

### Watch this Video for More Reference ### 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: Solve $\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{\sin x-x}{{{x}^{3}}}$

Solution: ### 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