NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives Exercise 12.1

NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.1 Limits and Derivatives help students understand the basic concept of limits, which is the foundation of calculus. This exercise introduces students to evaluating limits using simple substitution and algebraic methods.

Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 12.1 focuses on finding the limit of a function as a variable approaches a particular value. It helps students develop the initial understanding required for derivatives and advanced calculus topics.

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives Exercise 12.1

Limits and Derivatives

Difficult

Q.

Evaluate the following limits in Exercises 1 to 22.

1. $\underset{\mathbf{x}\mathbf{\to }\mathbf{3}}{\mathbf{lim}}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{)}$

2. $\underset{\mathbf{x}\mathbf{\to }\mathbf{\pi }}{\mathbf{lim}}\mathbf{(}\mathbf{x}\mathbf{-}\frac{22}{7}\mathbf{)}$

3. $\underset{\mathbf{r}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}{\mathbf{\pi r}}^{\mathbf{2}}$

4. $\underset{\mathbf{x}\mathbf{\to }\mathbf{4}}{\mathbf{lim}}\mathbf{\left(}\frac{\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}}{\mathbf{x}\mathbf{-}\mathbf{2}}\mathbf{\right)}$

5. $\underset{\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{1}}{\mathbf{lim}}\mathbf{\left(}\frac{{\mathbf{x}}^{\mathbf{10}}\mathbf{+}{\mathbf{x}}^{\mathbf{5}}\mathbf{+}\mathbf{1}}{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{\right)}$

6. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}}^5\mathbf{-}\mathbf{1}}{x}$

7. $\underset{\mathbf{x}\mathbf{\to }\mathbf{2}}{\mathbf{lim}}\mathbf{\left(}\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{10}}{{\mathbf{x}}^2\mathbf{-}\mathbf{4}}\mathbf{\right)}$

8. $\underset{\mathbf{x}\mathbf{\to }\mathbf{3}}{\mathbf{lim}}\frac{{\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{81}}{\mathbf{2}{\mathbf{x}}^2\mathbf{-}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{3}}$

9. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\left(}\frac{\mathbf{ax}\mathbf{+}\mathbf{b}}{\mathbf{cx}\mathbf{+}\mathbf{1}}\mathbf{\right)}$

10. $\underset{\mathbf{z}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\frac{{\mathbf{z}}^{\frac{1}{3}}\mathbf{-}\mathbf{1}}{{\mathbf{z}}^{\frac{1}{6}}\mathbf{-}\mathbf{1}}$

11. $\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{\left(}\frac{{\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}}{{\mathbf{cx}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{a}}\mathbf{\right)}\mathbf{,}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{\ne }\mathbf{0}$

12. $\underset{\mathbf{x}\mathbf{\to }\mathbf{-}\mathbf{2}}{\mathbf{lim}}\frac{\frac{1}{x}\mathbf{+}\frac{\mathbf{1}}{2}}{\mathbf{x}\mathbf{+}\mathbf{2}}$

13. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathrm{sinax}}{\mathrm{bx}}$

14. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{x}}{\mathbf{sin}\mathbf{ }\mathbf{b}\mathbf{ }\mathbf{x}}\mathbf{,}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\ne }\mathbf{0}$

15. $\underset{\mathbf{x}\mathbf{\to }\mathbf{\pi }}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{(}\mathbf{\pi }\mathbf{-}\mathbf{x}\mathbf{)}}{\mathbf{\pi }\mathbf{(}\mathbf{\pi }\mathbf{-}\mathbf{x}\mathbf{)}}$

16. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\cos x}{\mathbf{(}\mathbf{\pi }\mathbf{-}\mathbf{x}\mathbf{)}}$

17. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}}{\mathbf{cosx}\mathbf{-}\mathbf{1}}$

18. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{ax}\mathbf{+}\mathbf{x}\mathbf{ }\mathbf{cos}\mathbf{ }\mathbf{x}}{\mathbf{b}\mathbf{ }\mathbf{sin}\mathbf{ }\mathbf{x}}$

19. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{x}\mathbf{ }\mathbf{sec}\mathbf{ }\mathbf{x}$

20. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{ }\mathbf{ax}\mathbf{+}\mathbf{bx}}{\mathbf{ax}\mathbf{+}\mathbf{sin}\mathbf{ }\mathbf{bx}}\mathbf{,}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{\ne }\mathbf{0}$

21. $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{(}\mathbf{cosec}\mathbf{ }\mathbf{x}\mathbf{-}\mathbf{cot}\mathbf{x}\mathbf{)}$

22. $\underset{\mathbf{x}\mathbf{\to }\frac{\mathbf{\pi }}{2}}{\mathbf{lim}}\frac{\mathbf{tan}\mathbf{2}\mathbf{x}}{\mathbf{x}\mathbf{-}\frac{\mathbf{\pi }}{2}}$

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Limits and Derivatives

Easy

Q.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{4}\mathbf{x}}\mathbf{-}\mathbf{1}}{\mathbf{x}}$

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Limits and Derivatives

Easy

Q.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{log}}_{\mathbf{e}}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{)}}{\mathbf{x}}$

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Limits and Derivatives

Easy

Q.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{x}\mathbf{(}{\mathbf{e}}^{\mathbf{x}}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{1}\mathbf{-}\mathbf{cosx}}$

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Limits and Derivatives

Easy

Q.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{3}}}{\mathbf{x}\mathbf{-}\mathbf{3}}$

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Limits and Derivatives

Easy

Q.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{sinx}}\mathbf{-}\mathbf{1}}{\mathbf{x}}$

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Limits and Derivatives

Easy

Q.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{5}}}{\mathbf{x}\mathbf{-}\mathbf{5}}$

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Limits and Derivatives

Easy

Q.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{+}\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{2}}}{\mathbf{x}}$

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Limits and Derivatives

Medium

Q.

Find the derivative of the following functions:

(i) sin x cos x
(ii) sec x
(iii) 5sec x + 4cos x
(iv) cosec x
(v) 3cot x + 5cosec x
(vi) 5sin x − 6cos x + 7
(vii) 2 tan x − 7sec x

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Limits and Derivatives

Medium

Q.

$\mathbf{\text{Find }}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\text{ and }}\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{\text{\hspace{0.33em}where\hspace{0.33em}f }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}2x+3,x\le 0\\ 3(x+1),x>0\end{array}$

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Limits and Derivatives

Easy

Q.

$\begin{array}{l}\mathrm{If}\mathrm{the}\mathrm{function}\text{\hspace{0.33em}f}\left(\text{x}\right)\mathrm{satisfies}\underset{\text{x}\to 1}{\lim }\frac{\text{f}\left(\text{x}\right)-2}{{\text{x}}^2-1}=\text{π},\\ \mathrm{evaluate}\underset{\text{x}\to 1}{\lim }\text{\hspace{0.33em}f}\left(\text{x}\right).\end{array}$

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Limits and Derivatives

Difficult

Q.

$\begin{array}{l}\mathbf{If}\mathbf{ }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{+}\mathbf{1}\mathbf{,}& \mathbf{x}\mathbf{<}\mathbf{0}\\ \mathbf{0}\mathbf{,}& \mathbf{x}\mathbf{=}\mathbf{0}\\ \mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{-}\mathbf{1}\mathbf{,}& \mathbf{x}\mathbf{>}\mathbf{0}\end{array}\\ \mathbf{For}\mathbf{ }\mathbf{what}\mathbf{ }\mathbf{value}\mathbf{ }\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{does}\mathbf{ }\underset{\mathbf{x}\mathbf{\to }\mathbf{a}}{\mathbf{lim}}\mathbf{ }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{ }\mathbf{exists}\mathbf{?}\end{array}$

View Solution

Limits and Derivatives

Medium

Q.

Let a₁, a₂, …, a_n be fixed real numbers and define a function f(x) = (x – a₁) (x – a₂)… (x – a_n).

$\mathbf{What}\mathbf{ }\mathbf{is}\underset{\mathbf{x}\mathbf{\to }{\mathbf{a}}_{\mathbf{1}}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{?}\mathbf{ }\mathbf{For}\mathbf{ }\mathbf{some}\mathbf{ }\mathbf{a}\mathbf{\ne }{\mathbf{a}}_{\mathbf{1}}\mathbf{,}{\mathbf{a}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{..}\mathbf{,}{\mathbf{a}}_{\mathbf{n}}\mathbf{,}\mathbf{compute}\mathbf{ }\underset{\mathbf{x}\mathbf{\to }\mathbf{a}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}$

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Limits and Derivatives

Medium

Q.

$\mathbf{Find}\mathbf{ }\mathbf{\text{}}\underset{\mathbf{x}\mathbf{\to }\mathbf{5}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{where}\mathbf{ }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{-}\mathbf{5}$

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Limits and Derivatives

Medium

Q.

$\begin{array}{l}\mathbf{Suppose}\mathbf{ }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{a}\mathbf{+}\mathbf{bx}\mathbf{,}\mathbf{x}\mathbf{<}\mathbf{1}\\ \mathbf{4}\mathbf{,}\mathbf{x}\mathbf{=}\mathbf{1}\\ \mathbf{b}\mathbf{-}\mathbf{ax}\mathbf{,}\mathbf{x}\mathbf{>}\mathbf{1}\end{array}\\ \mathbf{and}\mathbf{ }\mathbf{if}\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{ }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{ }\mathbf{what}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{possible}\mathbf{ }\mathbf{values}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{b}\mathbf{?}\end{array}$

View Solution

Limits and Derivatives

Easy

Q.

$\mathbf{Find}\underset{\mathbf{\text{x}}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\mathbf{\text{\hspace{0.33em}f}}\left(\text{x}\right)\mathbf{,}\mathbf{ }\mathbf{where}\mathbf{ }\mathbf{\text{f}}\left(\text{x}\right)\mathbf{=}\mathbf{\{}\begin{array}{lll}\frac{\mathbf{\text{x}}}{\left|\text{x}\right|}\mathbf{,}& & \mathbf{\text{x}}\mathbf{\ne }\mathbf{0}\\ \mathbf{0}\mathbf{,}& & \mathbf{\text{x}}\mathbf{=}\mathbf{0}\end{array}$

View Solution

Limits and Derivatives

Easy

Q.

$\mathbf{Find}\mathbf{ }\underset{\mathbf{x}\mathbf{\to }\mathbf{1}}{\mathbf{lim}}\mathbf{ }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{where}\mathbf{ }\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{\le }\mathbf{1}\\ \mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{>}\mathbf{1}\end{array}$

View Solution

Limits and Derivatives

Easy

Q.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{log}\mathbf{(}\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{3}}\mathbf{)}}{{\mathbf{sin}}^{\mathbf{3}}\mathbf{x}}$

View Solution

The solutions are explained in a clear, step-by-step format so students can easily understand how to evaluate limits and avoid common mistakes in exams.

Question 1

Evaluate:
lim (x → 3) (x + 3)

Solution:
Substitute x = 3
= 3 + 3
= 6

Question 2

Evaluate:
lim (x → π) (x − 22/7)

Solution:
= π − 22/7

Question 3

Evaluate:
lim (r → 1) πr²

Solution:
= π(1)²
= π

Question 4

Evaluate:
lim (x → 4) (4x + 3)/(x − 2)

Solution:
= (4×4 + 3)/(4 − 2)
= (16 + 3)/2
= 19/2

Question 5

Evaluate:
lim (x → −1) (x¹⁰ + x⁵ + 1)/(x − 1)

Solution:
Substitute x = −1
Numerator = 1 − 1 + 1 = 1
Denominator = −2
= −1/2

Question 6

Evaluate:
lim (x → 0) [(1 + x)⁵ − 1]/x

Solution:
Using standard result:
lim (x → 0) [(1 + x)ⁿ − 1]/x = n

So,
= 5

Question 7

Evaluate:
lim (x → 2) (3x² − x − 10)/(x² − 4)

Solution:
Substitute x = 2 → 0/0 form

Factorize:
3x² − x − 10 = (x − 2)(3x + 5)
x² − 4 = (x − 2)(x + 2)

Cancel (x − 2):
= (3x + 5)/(x + 2)

Put x = 2:
= (6 + 5)/4
= 11/4

Question 8

Evaluate:
lim (x → 3) (x⁴ − 81)/(2x² − 5x − 3)

Solution:
x⁴ − 81 = (x − 3)(x + 3)(x² + 9)
2x² − 5x − 3 = (x − 3)(2x + 1)

Cancel (x − 3):
= (x + 3)(x² + 9)/(2x + 1)

Put x = 3:
= (6 × 18)/7
= 108/7

Question 9

Evaluate:
lim (x → 0) (ax + b)/(cx + 1)

Solution:
= b/1
= b

Question 10

Evaluate:
lim (z → 1) (z^(1/3) − 1)/(z^(1/6) − 1)

Solution:
Let z = t⁶

Then expression becomes:
(t² − 1)/(t − 1)

= (t − 1)(t + 1)/(t − 1)
= t + 1

Put t = 1
= 2

Question 11

Evaluate:
lim (x → 1) (ax² + bx + c)/(cx² + bx + a)

Solution:
Substitute x = 1

Numerator = a + b + c
Denominator = c + b + a

= same

= 1

Question 12

Evaluate:
lim (x → −2) [(1/x + 1/2)/(x + 2)]

Solution:
(1/x + 1/2) = (x + 2)/(2x)

So expression becomes:
[(x + 2)/(2x)] ÷ (x + 2)

Cancel (x + 2):
= 1/(2x)

Put x = −2
= −1/4

Question 13

Evaluate:
lim (x → 0) sin(ax)/(bx)

Solution:
= (a/b) × [sin(ax)/(ax)]
= (a/b) × 1
= a/b

Question 14

Evaluate:
lim (x → 0) sin(ax)/sin(bx)

Solution:
= a/b

Question 15

Evaluate:
lim (x → π) sin(π − x)/[π(π − x)]

Solution:
sin(π − x) = sin x

Let y = π − x

= sin y / (πy)
= 1/π

Question 16

Evaluate:
lim (x → 0) cos x/(π − x)

Solution:
= 1/π

Question 17

Evaluate:
lim (x → 0) (cos 2x − 1)/(cos x − 1)

Solution:
Using identities:

cos 2x − 1 = −2 sin²x
cos x − 1 = −2 sin²(x/2)

After simplification:
= 4

Question 18

Evaluate:
lim (x → 0) (ax + x cos x)/(b sin x)

Solution:
= (a + 1)/b

Question 19

Evaluate:
lim (x → 0) x sec x

Solution:
= x / cos x
= 0

Question 20

Evaluate:
lim (x → 0) (sin(ax) + bx)/(ax + sin(bx))

Solution:
= 1

Question 21

Evaluate:
lim (x → 0) (csc x − cot x)

Solution:
= 0

Question 22

Evaluate:
lim (x → π/2) tan(2x)/(x − π/2)

Solution:
= 2

Question 23

Piecewise function

Answer:
lim (x → 0) = 3
lim (x → 1) = 6

Question 24, 25, 26

Answer:
Limit does not exist

Question 27

Answer:
= 0

Question 28

Answer:
a = 0, b = 4

Question 29

Answer:
= 0

Question 30

Answer:
Limit exists for all a ≠ 0

Question 31

Answer:
lim (x → 1) = 2

Question 32

Answer:
Limit exists when m = n

FAQs – Class 11 Maths Chapter 12 Exercise 12.1 Limits and Derivatives

Q1. What is the focus of Exercise 12.1?
Exercise 12.1 focuses on understanding and evaluating the limits of functions using basic methods.

Q2. What is a limit?
A limit is the value that a function approaches as the input (x) approaches a particular value.

Q3. Why is Exercise 12.1 important for exams?
This exercise builds the foundation for derivatives and calculus, which are important topics in higher classes and competitive exams.

Q4. How can students prepare effectively for Exercise 12.1?
Students should practice evaluating limits using substitution, simplify algebraic expressions carefully, and solve different types of limit problems.