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Updated 6 August 2026

NCERT Solutions

Limits and Derivatives

Madhya Pradesh Board · Class 11 · Mathematics

NCERT Solutions for Limits and Derivatives — Madhya Pradesh Board Class 11 Mathematics.

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EXERCISE 12.1

\lim_{x \to 3} x + 3

Show solution

Substitute

x=3

since this is a polynomial expression:

\lim_{x\to 3}(x+3)=3+3=6.

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\lim_{x \to \pi} \left( x - \frac{22}{7} \right)

Show solution

Since the expression is continuous, substitute

x=\pi

\lim_{x\to \pi}\left(x-\frac{22}{7}\right)=\pi-\frac{22}{7}.

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\lim_{r \to 1} \pi r^2

Show solution

Use direct substitution:

\lim_{r\to 1}\pi r^2=\pi(1)^2=\pi.

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\lim_{x \to 4} \frac{4x + 3}{x - 2}

Show solution

Substitute

x=4

\lim_{x\to 4}\frac{4x+3}{x-2}=\frac{4(4)+3}{4-2}=\frac{16+3}{2}=\frac{19}{2}.

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\lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1}

Show solution

Substitute

x=-1

\lim_{x\to -1}\frac{x^{10}+x^5+1}{x-1} =\frac{(-1)^{10}+(-1)^5+1}{-1-1} =\frac{1-1+1}{-2} = -\frac12.

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\lim_{x \to 0} \frac{(x + 1)^5 - 1}{x}

Show solution

Use the standard limit for

(x+1)^5

by expansion or direct simplification:

\frac{(x+1)^5-1}{x} =\frac{1+5x+\cdots-1}{x}.

The leading term is

5x

, so the limit is

\lim_{x\to 0}\frac{(x+1)^5-1}{x}=5.

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\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}

Show solution

Factor the numerator and denominator:

3x^2-x-10=(3x+5)(x-2), \qquad x^2-4=(x-2)(x+2).

\lim_{x\to 2}\frac{3x^2-x-10}{x^2-4} =\lim_{x\to 2}\frac{(3x+5)(x-2)}{(x-2)(x+2)} =\lim_{x\to 2}\frac{3x+5}{x+2}.

Now substitute

x=2

\frac{3(2)+5}{2+2}=\frac{11}{4}.

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\lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3}

Show solution

Factor both polynomials:

x^4-81=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9),

2x^2-5x-3=(2x+1)(x-3).

Cancel

(x-3)

\lim_{x\to 3}\frac{x^4-81}{2x^2-5x-3} =\lim_{x\to 3}\frac{(x+3)(x^2+9)}{2x+1}.

Substitute

x=3

\frac{(6)(18)}{7}=\frac{108}{7}.

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\lim_{x \to 0} \frac{ax + b}{cx + 1}

Show solution

x=0

\lim_{x\to 0}\frac{ax+b}{cx+1}=\frac{a(0)+b}{c(0)+1}=b.

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\lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}

Show solution

Let

y=z^{1/6}

. Then

z^{1/3}=y^2

and as

z\to 1

y\to 1

.

So

\lim_{z\to 1}\frac{z^{1/3}-1}{z^{1/6}-1} =\lim_{y\to 1}\frac{y^2-1}{y-1} =\lim_{y\to 1}(y+1)=2.

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\lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a}, a + b + c \neq 0

Show solution

x=1

\frac{ax^2+bx+c}{cx^2+bx+a} =\frac{a+b+c}{c+b+a}=1,

provided

a+b+c\ne 0

so the expression is defined. Hence the limit is

1

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\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}

Show solution

Simplify first:

\frac{\frac1x+\frac12}{x+2}= \frac{\frac{2+x}{2x}}{x+2}=\frac{2+x}{2x(x+2)}.

Now substitute

x=-2

by cancelling the common factor

(x+2)

from the original form:

\frac{1}{x}+\frac12=\frac{2+x}{2x}.

So the expression becomes

\frac{\frac{2+x}{2x}}{x+2}=\frac{1}{2x}.

Thus

\lim_{x\to -2}\frac{\frac1x+\frac12}{x+2}=\frac{1}{2(-2)}=-\frac14.

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\lim_{x \to 0} \frac{\sin ax}{bx}

Show solution

Use the standard limit

\lim_{u\to 0}\frac{\sin u}{u}=1

\lim_{x\to 0}\frac{\sin ax}{bx} =\frac{a}{b}\lim_{x\to 0}\frac{\sin ax}{ax} =\frac{a}{b}\cdot 1 =\frac{a}{b}.

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\lim_{x \to 0} \frac{\sin ax}{\sin bx}, a, b \neq 0

Show solution

Write

\frac{\sin ax}{\sin bx}=\frac{\sin ax}{ax}\cdot \frac{bx}{\sin bx}\cdot \frac{a}{b}.

x\to 0

, each of the first two factors tends to

1

. Therefore,

\lim_{x\to 0}\frac{\sin ax}{\sin bx}=\frac{a}{b}.

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\lim_{x\to \pi}\frac{\sin(\pi - x)}{\pi(\pi - x)}

Show solution

Since

\sin(\pi-x)=\sin x

and

\pi(\pi-x)\to \pi\cdot 0=0

, but we use the standard form:

\lim_{x\to \pi}\frac{\sin(\pi-x)}{\pi(\pi-x)} =\frac{1}{\pi}\lim_{x\to \pi}\frac{\sin(\pi-x)}{\pi-x}.

Let

u=\pi-x

. Then

u\to 0

and

\frac{1}{\pi}\lim_{u\to 0}\frac{\sin u}{u}=\frac{1}{\pi}\cdot 1=\frac{1}{\pi}.

So the limit is

\frac{1}{\pi}

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\lim_{x\to 0}\frac{\cos x}{\pi - x}

Show solution

Direct substitution gives

\lim_{x\to 0}\frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}.

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\lim_{x\to 0}\frac{\cos 2x - 1}{\cos x - 1}

Show solution

Use identities:

\cos 2x-1 = -2\sin^2 x, \qquad \cos x-1 = -2\sin^2\frac{x}{2}.

Then

\frac{\cos 2x-1}{\cos x-1}= \frac{-2\sin^2 x}{-2\sin^2(x/2)} =\left(\frac{\sin x}{\sin(x/2)}\right)^2.

Now

\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2},

\left(\frac{\sin x}{\sin(x/2)}\right)^2 = (2\cos(x/2))^2.

x\to 0

, this tends to

4

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\lim_{x\to 0}\frac{ax + x\cos x}{b\sin x}

Show solution

Factor

x

from the numerator:

\frac{ax+x\cos x}{b\sin x}=\frac{x(a+\cos x)}{b\sin x}.

Now use

\lim_{x\to 0}\frac{x}{\sin x}=1

and

\cos 0=1

\lim_{x\to 0}\frac{x(a+\cos x)}{b\sin x}=\frac{a+1}{b}.

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\lim_{x\to 0}x\sec x

Show solution

Since

\sec x

is continuous at

0

and

\sec 0=1

\lim_{x\to 0} x\sec x = \left(\lim_{x\to 0}x\right)\left(\lim_{x\to 0}\sec x\right)=0\cdot 1=0.

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\lim_{x\to 0}\frac{\sin ax + bx}{ax + \sin bx} a,b,a + b\neq 0,

Show solution

Use

\sin kx \sim kx

x\to 0

\lim_{x\to 0}\frac{\sin ax + bx}{ax + \sin bx} =\lim_{x\to 0}\frac{ax+bx}{ax+bx}=1,

provided

a+b\neq 0

so the cancellation is valid in the limit.

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\lim_{x\to 0}(\operatorname{cosec}x - \cot x)

Show solution

Use the standard identity

\csc x-\cot x=\frac{1-\cos x}{\sin x}.

Then

\lim_{x\to 0}(\csc x-\cot x)=\lim_{x\to 0}\frac{1-\cos x}{\sin x}.

Now

1-\cos x\to 0 \quad\text{and}\quad \sin x\sim x,

so the limit is

0

. More directly, using

1-\cos x=2\sin^2(x/2)

\frac{1-\cos x}{\sin x}=\frac{2\sin^2(x/2)}{\sin x}\to 0.

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\lim_{x\to \frac{\pi}{2}}\frac{\tan 2x}{x - \frac{\pi}{2}}

Show solution

Let

u=x-\frac{\pi}{2}

, so

u\to 0

. Then

\tan 2x = \tan\left(\pi+2u\right)=\tan(2u),

since tangent has period

\pi

. Therefore

\lim_{x\to \frac{\pi}{2}}\frac{\tan 2x}{x-\frac{\pi}{2}} =\lim_{u\to 0}\frac{\tan(2u)}{u} =2\lim_{u\to 0}\frac{\tan(2u)}{2u}=2\cdot 1=2.

But because

\tan(\pi+2u)=\tan(2u)

, the computation gives

2

, not

-4

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23Find

\lim_{x\to 0}f(x)

and

\lim_{x\to 1}f(x)

, where

f(x) = \begin{cases} 2x + 3, & x\leq 0 \\ 3(x + 1), & x > 0 \end{cases}

Show solution

For

f(x)=\begin{cases}2x+3,&x\le 0\\3(x+1),&x>0\end{cases}

x\to 0

:
- left side gives

2(0)+3=3

- right side gives

3(0+1)=3

\lim_{x\to 0}f(x)=3.

x\to 1

, since

1>0

, use

f(x)=3(x+1)

\lim_{x\to 1}f(x)=3(1+1)=6.

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24Find

\lim_{x\to 1}f(x)

, where

f(x) = \begin{cases} x^2 - 1, & x \leq 1 \\ -x^2 - 1, & x > 1 \end{cases}

Show solution

x=1

,
- from the left:

x^2-1\to 1-1=0

- from the right:

-x^2-1\to -1-1=-2

The left and right limits are different, so the limit does not exist.

But the question asks for

\lim_{x\to 1}f(x)

, and since the one-sided limits are unequal, the limit does not exist.

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25Evaluate

\lim_{x\to 0}f(x)

, where

f(x) = \begin{cases} \frac{|x|}{x}, & x\neq 0 \\ 0, & x = 0 \end{cases}

Show solution

For

x\ne 0

f(x)=\frac{|x|}{x}= \begin{cases} 1,&x>0,\\ -1,&x<0. \end{cases}

\lim_{x\to 0^+}f(x)=1,\qquad \lim_{x\to 0^-}f(x)=-1.

Since the left and right limits are not equal, the limit at

0

does not exist.

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26Find

\lim_{x\to 0}f(x)

, where

f(x) = \begin{cases} \frac{x}{|x|}, & x\neq 0 \\ 0, & x = 0 \end{cases}

Show solution

For

x\neq 0

, we have

- if

x>0

, then

|x|=x

, so

\dfrac{x}{|x|}=1

- if

x<0

, then

|x|=-x

, so

\dfrac{x}{|x|}=-1

Thus the left hand limit at

0

-1

and the right hand limit is

1

. Since they are not equal, the limit at

x=0

does not exist.

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27Find

\lim_{x\to 5}f(x)

, where

f(x) = |x| - 5

Show solution

Since

|x|

is continuous, as

x\to 5

we get

\lim_{x\to 5} (|x|-5)=|5|-5=5-5=0.

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28Suppose

f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases}

and if

\lim_{x\to 1}f(x) = f(1)

what are possible values of

a

and

b

?Show solution

For the limit to equal

f(1)=4

, the left hand limit and right hand limit at

x=1

must both be 4.

For

x<1

\lim_{x\to 1^-}f(x)=a+b(1)=a+b

So,

a+b=4

For

x>1

\lim_{x\to 1^+}f(x)=b-a(1)=b-a

So,

b-a=4

Solving:

(a+b)+(b-a)=4+4 \Rightarrow 2b=8 \Rightarrow b=4

Then

a+b=4 \Rightarrow a+4=4 \Rightarrow a=0

So the possible values are **

a=0

and

b=4

**.

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29Let

a_1, a_2, \ldots, a_n

be fixed real numbers and define a function

f(x) = (x - a_1)(x - a_2)\ldots(x - a_n).

What is

\lim_{x \to a_i} f(x)

? For some

a \neq a_1, a_2, \ldots, a_n

, compute

\lim_{x \to a} f(x)

.Show solution

x\to a_i

for any one of the fixed numbers

a_i

, then one factor becomes

0

\lim_{x\to a_i}(x-a_1)(x-a_2)\cdots(x-a_n)=0.

a\neq a_1,a_2,\ldots,a_n

, then none of the factors is zero at

x=a

, so the product is a polynomial and its limit equals its value at

a

\lim_{x\to a}f(x)=f(a)=(a-a_1)(a-a_2)\cdots(a-a_n).

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30If

f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases}

.Show solution

For

x<0

|x|=-x

, so

f(x)=|x|+1=-x+1.

Hence as

x\to 0^-

f(x)\to 1.

For

x>0

|x|=x

, so

f(x)=|x|-1=x-1.

Hence as

x\to 0^+

f(x)\to -1.

Since the left and right hand limits are different, the limit at

x=0

does not exist.

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31If the function

f(x)

satisfies

\lim_{x \to 1} \frac{f(x) - 2}{x^2 - 1} = \pi

, evaluate

\lim_{x \to 1} f(x)

.Show solution

Given

\lim_{x\to 1}\frac{f(x)-2}{x^2-1}=\pi.

x\to 1

, we have

x^2-1\to 0

. For the quotient to have a finite limit, the numerator must also approach

0

\lim_{x\to 1}(f(x)-2)=0.

Therefore

\lim_{x\to 1} f(x)=2.

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32If

f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}

. For what integers

m

and

n

does both

\lim_{x \to 0} f(x)

Show solution

We need both limits to exist.

At

x\to 0

:
- from

x<0

f(x)=mx^2+n \to n

- from

0\le x\le 1

f(x)=nx+m \to m

So for

\lim_{x\to 0}f(x)

to exist,

n=m.

x\to 1

:
- from

0\le x\le 1

f(x)=nx+m \to n+m

- from

x>1

f(x)=nx^3+m \to n+m

These are automatically equal, so the limit at

x=1

exists for any integers once

m=n

.

Hence the required condition is

\boxed{m=n}

for integers

m,n

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EXERCISE 12.2

1Find the derivative of

x^2 - 2

x = 10

.Show solution

Let

f(x)=x^2-2

. Its derivative is

f'(x)=2x.

x=10

f'(10)=2\cdot 10=20.

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2Find the derivative of

x

x = 1

.Show solution

For

f(x)=x

, the derivative is

f'(x)=1.

So at

x=1

it is still

1

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3Find the derivative of

99x

x = 100

.Show solution

For

f(x)=99x

, the derivative is the constant

f'(x)=99.

So at

x=100

, the derivative is

99

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4(i)Find the derivative of the following functions from first principle.Show solution

Using the first principle results from the chapter:

1. For

f(x)=x^3-27

\frac{d}{dx}(x^3-27)=3x^2.

2. For

f(x)=(x-1)(x-2)=x^2-3x+2

\frac{d}{dx}(x^2-3x+2)=2x-3.

3. For

f(x)=\frac{1}{x^2}=x^{-2}

\frac{d}{dx}(x^{-2})=-2x^{-3}=-\frac{2}{x^3}.

4. For

f(x)=\frac{x+1}{x-1}

, using the quotient rule,

\frac{d}{dx}\left(\frac{x+1}{x-1}\right)=\frac{(1)(x-1)-(x+1)(1)}{(x-1)^2} =\frac{x-1-x-1}{(x-1)^2} =-\frac{2}{(x-1)^2}.

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4(ii)Find the derivative of the following functions from first principle.Show solution

For

f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdots+\frac{x^2}{2}+x+1,

we differentiate term by term:

f'(x)=x^{99}+x^{98}+\cdots+x+1.

Now,

f'(1)=1+1+\cdots+1=100

and

f'(0)=0+0+\cdots+0+1=1.

Hence

f'(1)=100f'(0).

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4(iii)Find the derivative of the following functions from first principle.Show solution

Using standard differentiation rules:

(i)

\frac{d}{dx}[(x-a)(x-b)] = (x-a)'(x-b)+(x-a)(x-b)'

=1\cdot(x-b)+(x-a)\cdot 1=2x-a-b.

(ii)

\frac{d}{dx}[(ax+b)(cx+d)^2]

Use product rule:

=ax+b \Rightarrow (ax+b)'=a, \quad (cx+d)^2 \Rightarrow 2(cx+d)\cdot c.

= a(cx+d)^2 + (ax+b)\cdot 2c(cx+d)

=(cx+d)\big(a(cx+d)+2c(ax+b)\big).

(iii)

\frac{d}{dx}\left(\frac{x-a}{x-b}\right) =\frac{(1)(x-b)-(x-a)(1)}{(x-b)^2} =\frac{x-b-x+a}{(x-b)^2} =\frac{a-b}{(x-b)^2}.

So the derivative is

\frac{a-b}{(x-b)^2}.

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4(iv)Find the derivative of the following functions from first principle.Show solution

Let

f(x)=\frac{x^n-a^n}{x-a}.

Using the quotient rule with numerator

u=x^n-a^n

and denominator

v=x-a

u'=nx^{n-1},\quad v'=1.

f'(x)=\frac{u'v-uv'}{v^2} =\frac{nx^{n-1}(x-a)-(x^n-a^n)}{(x-a)^2}.

Expanding the numerator:

=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2} =\frac{(n-1)x^n-nax^{n-1}+a^n}{(x-a)^2}.

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5For the function

f (x) = \frac {x ^ {1 0 0}}{1 0 0} + \frac {x ^ {9 9}}{9 9} + \dots + \frac {x ^ {2}}{2} + x + 1.

Show solution

For

f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\cdots+\frac{x^2}{2}+x+1,

differentiate term by term:

f'(x)=x^{99}+x^{98}+\cdots+x+1.

This is the required derivative.

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6Find the derivative of

x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n

for some fixed real number

a

7(i)For some constants

a

and

b

, find the derivative of

7(ii)For some constants

a

and

b

, find the derivative of

7(iii)For some constants

a

and

b

, find the derivative of

8Find the derivative of

\frac{x^n - a^n}{x-a}

for some constant

a

9(i)Find the derivative of

9(vi)Find the derivative of

10Find the derivative of

\cos x

from first principle.

11(ii)Find the derivative of the following functions:

11(v)Find the derivative of the following functions:

11(vii)Find the derivative of the following functions:

EXERCISE 12.1

(x + a)

(px + q)\left(\frac{r}{x} + s\right)

(ax + b)(cx + d)^2

\frac{ax + b}{cx + d}

\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}

\frac{1}{ax^2 + bx + c}

\frac{ax + b}{px^2 + qx + r}

\frac{px^2 + qx + r}{ax + b}

\frac{a}{x^4} - \frac{b}{x^2} + \cos x

4\sqrt{x} - 2

(ax + b)^n

(ax + b)^n (cx + d)^m

\sin(x + a)

\csc x \cot x

\frac{\cos x}{1 + \sin x}

\frac{\sin x + \cos x}{\sin x - \cos x}

\frac{\sec x - 1}{\sec x + 1}

\sin^n x

\frac{a + b \sin x}{c + d \cos x}

\frac{\sin(x + a)}{\cos x}

x^4 (5 \sin x - 3 \cos x)

(x^2 + 1) \cos x

(ax^2 + \sin x)(p + q \cos x)

(x + \cos x)(x - \tan x)

\frac{4x + 5 \sin x}{3x + 7 \cos x}

\frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x}

\frac{x}{1 + \tan x}

(x + \sec x)(x - \tan x)

\frac{x}{\sin^n x}

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Frequently Asked Questions

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Limits and Derivatives covers several key topics that are frequently asked in Madhya Pradesh Board Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.

How to score full marks in Limits and Derivatives — Madhya Pradesh Board Class 11 Mathematics?

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Limits and Derivatives

Studying Limits and Derivatives? Get the full interactive chapter.

EXERCISE 12.1

EXERCISE 12.2

EXERCISE 12.1

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