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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

\(\begin{array}{l}\underset{x\to c}{\mathop{\lim }}\,\,f(x)=L\end{array} \)
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)

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

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

Examples

    • \(\begin{array}{l}\underset{x\to 1}{\mathop{\lim }}\,\,{{x}^{2}}+1={{1}^{2}}+1=2\end{array} \)
    • \(\begin{array}{l}\underset{x\to 0}{\mathop{\lim }}\,\,{{x}^{2}}-x={{0}^{2}}-0=0\end{array} \)
    • \(\begin{array}{l}\underset{x\to 2}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4}{x+3}=\frac{4-4}{2+3}=0\end{array} \)

(c)In Limits, we have in determinant forms such as

\(\begin{array}{l}\frac{0}{0},\frac{\infty }{\infty },0\times \infty ,\infty \times \infty ,{{1}^{\infty }},0{}^\circ ,\infty {}^\circ\end{array} \)

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

Watch this Video for More Reference

Calculus Problems

Calculus Problems

Expected Questions and Solutions

Limit Continuity and Differentiability

Use of Expansion in Evaluating limits

Some Important Expansions

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

Some Important limits

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

Example: Solve

\(\begin{array}{l}\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{\sin x-x}{{{x}^{3}}}\end{array} \)

Solution:

Evaluation of Algebric Limits

Direct substitution method

Example:

\(\begin{array}{l}\underset{x\to 1}{\mathop{\lim }}\,\,(3{{x}^{2}}+4x+5)\end{array} \)
= 3(1)2 + 4(1) + 5 = 12

Example:

\(\begin{array}{l}\underset{x\to 2}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4}{x+3}=\frac{4-4}{2+3}=\frac{0}{5}=0\end{array} \)

Factorization method

Example:

\(\begin{array}{l}\underset{x\to 2}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-5x+6}{{{x}^{2}}-4}\end{array} \)

Solution:

\(\begin{array}{l}\underset{x\to 2}{\mathop{\lim }}\,\,\frac{(x-2)(x-3)}{(x+2)(x-2)}\Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\frac{x-3}{x+2}\end{array} \)
=
\(\begin{array}{l}\frac{-1}{4}\end{array} \)

Rationalization method

\(\begin{array}{l}\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sqrt{2+x}-\sqrt{2}}{x}\end{array} \)

Solution:

\(\begin{array}{l}\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\left( \sqrt{2+x}-\sqrt{2} \right)\left( \sqrt{2+x}+\sqrt{2} \right)}{x\left( \sqrt{2+x}+\sqrt{2} \right)}\end{array} \)

=

\(\begin{array}{l}\frac{1}{2\sqrt{2}}\end{array} \)

 

Using Result:

\(\begin{array}{l}\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}\end{array} \)

Example:

\(\begin{array}{l}\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{10}}-{{2}^{10}}}{{{x}^{5}}-{{2}^{5}}}\end{array} \)

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

\(\begin{array}{l}\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=f(a)\end{array} \)

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

Else, a function f(x) is said to be discontinuous Function.

Example 1: f(x) =

\(\begin{array}{l}\frac{1}{2\sin x-1},\end{array} \)
Discuss the continuity or discontinuity.

Solution: Clearly the function will be not defined for

\(\begin{array}{l}\sin x=\frac{1}{2}=\sin \frac{\pi }{6}\end{array} \)

Function is discontinuous for

\(\begin{array}{l}x=n\,\pi +{{(-1)}^{n}}\frac{\pi }{6}\end{array} \)

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

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

Solution: =

\(\begin{array}{l}f(4)=\underset{x\to 4}{\mathop{\lim }}\,\,\,\,\,\frac{{{x}^{4}}-256}{x-4}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{4}}-{{4}^{4}}}{x-4}={{4.4}^{4-1}}=256\end{array} \)

Example 3: Discuss the continuity of

(a) Sgn (x3x)

(b) f(x) =

\(\begin{array}{l}\left[ \frac{2}{1+{{x}^{2}}} \right]\end{array} \)
, x > 0 [ ] Solution:
(a) f(x) = sgn (x3–x)

Here x3 – x = 0, so, x =0, –1, 1

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

(b)

\(\begin{array}{l}\frac{2}{1+{{x}^{2}}}\end{array} \)
, x > 0 is a monotonically decimal function

Hence f(x) =

\(\begin{array}{l}\left[ \frac{2}{1+{{x}^{2}}} \right],\,\,\,x\ge 0\end{array} \)
is discontinuous,

When

\(\begin{array}{l}\frac{2}{1+{{x}^{2}}}\end{array} \)
is on integer

\(\begin{array}{l}\Rightarrow \frac{2}{1+{{x}^{2}}}=1,\,2\,\,\,\,\,\,at\,\,\,\,x=1,0\end{array} \)

Example 4: Discuss the continuity of f(x) =

\(\begin{array}{l}\left\{\begin{matrix} x-2 & x\leq 0\\ 4-x^2 &x>0 \end{matrix}\right.\end{array} \)
at x = 0

Solution:

\(\begin{array}{l}\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,\,(x-2)=-2\end{array} \)

Intermediate Value Theorem

If f is continuous on [a, b] and f(a) ≠ f(b), then for any value

\(\begin{array}{l}c\in (f(a),f(b))\end{array} \)
, there is at least one number in x0 (a, b) for which f(x0) = c

Limits, Continuity and Differentiability JEE Main Questions

12 Must-Do JEE Questions of Limits Continuity and Differentiability

Differentiability

A function, say f(x) is said to be differentiable at the point x = a if the derivative f ‘(a) exists at every point in its domain.

Existence of Derivative

Right and left hand derivative

F.H.D: f’(0+) =

\(\begin{array}{l}\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\end{array} \)
\(\begin{array}{l}\frac{f(a+h)-(a)}{h}\end{array} \)

L.H.D: F’

\(\begin{array}{l}({{a}^{-}})\end{array} \)
=
\(\begin{array}{l}\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\end{array} \)
\(\begin{array}{l}\frac{h(a-h)-f(a)}{-h}\end{array} \)

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.

Video Lesson:

Limits, Continuity and Differentiability – Part 1

Limits, Continuity and Differentiability

Limits, Continuity and Differentiability – Part 2

Limit Continuity and Differentiability

Limits, Continuity and Differentiability – Important Topics

Limit Continuity and Differentiability - Important Topics

Limits, Continuity and Differentiability – Important Questions

Limit Continuity and Differentiability - Important Questions

Limits, Continuity and Differentiability – Revision Lesson Part 1

Limits, Continuity and Differentiability – Revision Lesson Part 2

Frequently Asked Questions

Give the definition of continuous function in Mathematics.

A function is said to be continuous at a point x = a, if limx→a f(x) exists and limx→a f(x) = f(a).

What do you mean by differentiability?

A function f(x) is said to be differentiable at the point x = a if the derivative f’(a) exists at every point in its domain.

How do you check the continuity of a function f(x) at x = a?

We have to check the left hand limit and right-hand limit and the value of the function at a point x=a. If L.H.L = R.H.L = value of the function at x = a, then the function is continuous at x = a.

What do you mean by the second derivative of a function?

Second derivative is the derivative of the derivative of a function. We denote the second derivative of f(x) by f’’(x).

Is continuity and differentiability related?

Yes. If a function is differentiable at any point in its domain, it will be continuous at that point. Vice versa is not always true.

What is the value of limx→ 0(log (1+x)/x)?

The value of limx→ 0(log (1+x)/x) = 1.

What is the value of  limx→ 0 (ex-1)/x?

The value of limx→ 0 (ex-1)/x = 1.

Is a continuous function always differentiable?

A continuous function need not be differentiable. If a function f(x) is differentiable at a point a, then it is continuous at the point a. But converse is not true.