The concept of limits is essential for the advancement of analysis, as the convergence and divergence of infinite series depend upon it. Mathematicians have successfully employed the theory of limits to define continuity, differentiability and the definite integral in terms of the limit concept. In this section, you will gain an in-depth understanding of this concept with the help of solved examples.

What are Limits?

The expression (\underset{x\to c}{\mathop{\lim }},,f(x)=L) means that, as (x) gets closer and closer to (C), (f(x)) gets closer and closer to (L). In this case, we say that the limit of (f) as (x) approaches (C) is (L).

Neighbourhood of a Point:

Let a be a real number and h is very close to 0, then

Left hand limit will be obtained when x $\rightarrow$ a$^{-}$

Similarly, Right Hand limit will be obtained when x $\rightarrow$ a+ or x = a + h.

Related Concepts:

Functions and Its Types

Limits, Continuity and Differentiability

Differentiation

Applications of Derivatives

Existence of a Limit

The limit will exist if the following conditions are met:

(a) (\underset{x\to {{a}^{-}}}{\mathop{\lim }}f(x) = \underset{x\to {{a}^{+}}}{\mathop{\lim }}f(x) ) i.e. Left Hand Limit = Right Hand Limit

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

Examples:

• Example 1
• Example 2
• Example 3

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$\underset{x\to 1}{\mathop{\lim }},,{{x}^{2}}+1={{1}^{2}}+1=2$

(\underset{x\to 0}{\mathop{\lim }},,{{x}^{2}}-x \rightarrow 0)

$\underset{x\to 2}{\mathop{\lim }},,\frac{{{x}^{2}}-4}{x+3} = \frac{(x^2 - 4) \mid_{x=2}}{(x+3) \mid_{x=2}} = \frac{4-4}{2+3} = 0$

(c) In Limits, we have indeterminate forms such as \(\frac{0}{0},\frac{\infty }{\infty },0\times \infty ,\infty \times \infty ,{{1}^{\infty }},0{}^\circ ,\infty {}^\circ\)

In these cases, we attempt to reduce the problem into a valid function.

Watch [This Video] for More Reference

Calculus Problems

Expected Questions and Solutions

Evaluating Limits Through Expansion

Some Important Expansions

1. (\log(1+x) = x - \frac{{x}^2}{2} + \frac{{x}^3}{3} - \frac{{x}^4}{4} + \dots )

2. $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} - \frac{x^4}{4!} + \cdots$$

3. $$\begin{array}{l}{{a}^{x}}=1+x\log a+\frac{{{x}^{2}}}{2!}{(\log;a)}^{2}+\dots\end{array}$$

4. (\begin{array}{l}\sin x = x - \frac{{x^3}}{6} + \frac{{x^5}}{120} + \dots \end{array})

5. $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots$

6. (\displaystyle \tan x = x + \frac{{{x}^{3}}}{3} + \frac{2}{15}{{x}^{5}} + \cdots )

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. $$\underset{x\to 0}{\mathop{\lim }},,,,\frac{\log (1+x)}{x}=1$$

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

Given:

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Solution:

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Evaluating Algebraic Limits

Direct Substitution Method

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

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

Factorization Method

Answer: \begin{array}{l}\underset{x\to 2}{\mathop{\lim }},,\frac{{{x}^{2}}-5x+6}{{{x}^{2}}-4}\end{array} = \frac{2-5+6}{2-4} = \frac{3}{-2} = -\frac{3}{2}\end{array}

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

Rationalization Method

Example:

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(\underset{x\to 0}{\mathop{\lim }},,\frac{\sqrt{2+x}-\sqrt{2}}{x} = \underset{x\to 0}{\mathop{\lim }},,\frac{\frac{2+x-2}{\sqrt{2+x}+\sqrt{2}}}{x} = \underset{x\to 0}{\mathop{\lim }},,\frac{1}{\sqrt{2+x}+\sqrt{2}} = \frac{1}{2\sqrt{2}})

Solution: (\underset{x\to 0}{\mathop{\lim }},,\frac{2}{x})

\(\frac{1}{2\sqrt{2}}\)

Result:

Using Result: $$\underset{x\to a}{\mathop{\lim }},\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$$

Answer: (\underset{x\to 2}{\mathop{\lim }},\frac{{{x}^{10}}-{{2}^{10}}}{{{x}^{5}}-{{2}^{5}}})

Continuity

What is Continuity?

Continuity is the property of a function or a curve that it is unbroken and without abrupt changes. It is a concept used in calculus and other branches of mathematics.

A function is said to be discontinuous if small changes in the input do not result in small changes in the output. Otherwise, the function is said to be continuous.

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

for all values of ε > 0, there exists a δ > 0 such that if |x - a| < δ, then |f(x) - f(a)| < ε

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

i.e. $$L.H.L = R.H.L = f(a)$$

A function f(x) is said to be discontinuous if it is not continuous.

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

Solution: Clearly the function will not be 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) = Kx^2 - 6x + 9 has a minimum value of 3?

The value of K must be 2.

Yes, (f(x)) is continuous at x = 4.

Given:

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Solution:

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

Example 3: What are the implications of the continuity of

(a) $\mathrm{sgn}(x^3 - x)$

(b) (\begin{array}{l}f(x) = \left[ \frac{2}{1 + x^2} \right], \quad x > 0\end{array})

Given:

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Solution:

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f(x) = sgn(x^3 - x)

Here, $$x^3 - x = 0$$, so, $$x = 0, -1, 1$$

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

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

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

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

(\Rightarrow \frac{2}{1+{{x}^{2}}}=1 \quad \text{at} \quad x=1, 0)

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

The function (f(x)) is continuous at (x=0) since the left-hand limit and the right-hand limit both exist and are equal to 0.

Given:

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Solution:

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

Intermediate Value Theorem

If $f$ is continuous on $[a, b]$ and $f(a) \neq f(b)$, then for any value $c \in (f(a), f(b))$, there is at least one number $x_0 \in (a, b)$ such that $f(x_0) = c$.

JEE Main Questions on Limits, Continuity, and Differentiability

12 Essential JEE Questions on Limits, Continuity, and Differentiability

Differentiability

The 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.

Existence of Derivative

Right-Hand and Left-Hand Derivatives

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

L.H.D: F'\(\displaystyle F'(a^-) = \lim_{h \to 0^-} \frac{h(a-h)-f(a)}{-h}\)

What are the conditions for a function to be non-differentiable?

The function f(x) is said to be non-differentiable at x = a if it is not continuous or its derivative does not exist at x = a.

(a) Both R.H.D and L.H.D exist, but they are not equal.

Either R.H.D. or L.H.D. (or both) is not finite.

(c) Either R.H.D or L.H.D, or both, may not exist.

Video Lesson:

Limits, Continuity, and Differentiability - Part 1

Limits, Continuity and Differentiability - Part 2

Limits, Continuity, and Differentiability: Important Topics

Important Questions on Limits, Continuity and Differentiability

Revision Lesson Part 1: Limits, Continuity and Differentiability

Limits, Continuity, and Differentiability - Revision Lesson Part 2

Continuity of a Function

Theorems on Continuity

Differentiability and Its Conditions

Differentiability on an Interval

![Differentiability in an Interval]()

Frequently Asked Questions

A continuous function is a function for which, given any two points in its domain, all points between those two points are also in its domain.

If $\lim_{x \to a}f(x)$ exists and $\lim_{x \to a}f(x) = f(a)$, then a function is said to be continuous at a point $x = a$.

Differentiability is a concept in calculus that refers to the ability of a function to be differentiated, or have its rate of change calculated.

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.

What is the process for checking the continuity of a function f(x) at x = a?

If the Left Hand Limit (L.H.L) equals the Right Hand Limit (R.H.L) and is equal to the value of the function at x = a, then the function is continuous at x = a.

What is the definition of the second derivative of a function?

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

Yes, continuity and differentiability are related.

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

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

The value of $$\lim_{x\to 0}\frac{\log(1+x)}{x} = 1.$$

The value of $\lim_{x \to 0} \frac{e^x - 1}{x}$ is 1.

The limit as $x \rightarrow 0$ of $\frac{e^x - 1}{x}$ is 1.

Is a continuous function always differentiable?

No, a continuous function is not 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. However, the converse is not true.