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# NCERT Solutions for Class 11 Maths Chapter 12 - Limits and Derivatives Exercise 12.1

Last updated date: 11th Aug 2024
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## NCERT Solutions for Maths Exercise 12.1 Class 11 Chapter 12 - FREE PDF Download

Chapter 12 Limits and derivatives form the foundation of calculus is an essential branch of mathematics. NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.1 the concept of limits, focusing on understanding how a function behaves as the input approaches a particular point. This exercise is crucial as it sets the stage for more complex topics in calculus.

Table of Content
1. NCERT Solutions for Maths Exercise 12.1 Class 11 Chapter 12 - FREE PDF Download
2. Glance on NCERT Class 11 Maths Chapter 12 Exercise 12.1 Solutions  | Vedantu
3. Formulas Used in Class 11 Chapter 12 Exercise 12.1
4. Access NCERT Solutions for Maths Class 11 Chapter 12 - Limits and Derivatives
4.1Exercise 12.1
5. Class 11 Maths Chapter 12: Exercises Breakdown
6. CBSE Class 11 Maths Chapter 12 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Class 11 maths NCERT solutions chapter 12 exercise 12.1 is vital for solving advanced problems in calculus and other mathematical fields. Student can also find the latest CBSE Class 11 Maths Syllabus solved by Vedantu’s team of master teachers in detailed step-by-step manner to ensure complete understanding of the approach, hence get them full marks.

## Glance on NCERT Class 11 Maths Chapter 12 Exercise 12.1 Solutions  | Vedantu

• Class 11 Chapter 12 explains the foundational concepts of limits and derivatives essential for understanding calculus.

• Understanding limits helps in analysing the behaviour of functions near specific points, especially where they might not be explicitly defined. It covers techniques for calculating limits, such as direct substitution and the use of limit laws.

• It contains the formulas and techniques to calculate limits for different functions.

• Understanding these concepts provides a foundation for solving more complex problems in calculus and its applications in science and engineering.

• Exercise 12.1 contains 32 Questions to practise this concept.

## Formulas Used in Class 11 Chapter 12 Exercise 12.1

Limits Formula

To express a function's limit, we represent it as:

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

Left Hand and Right-Hand Limits

• $f(a-0) = \lim_{x\to a^{-}}f(x) = \lim_{h\to 0}f(a-h)$

• $f(a+0) = \lim_{x\to a{+}}f(x) = \lim_{h\to 0}f(a+h)$

Existence of Limit

$\lim_{x\to a}f(x)$ exists, if

(i) $\lim_{x\to a^{-}}f(x)$ and $\lim_{x\to a^{+}}f(x)$ both exists

(ii) $\lim_{x\to a^{-}}f(x) = \lim_{x\to a^{+}}f(x)$

Competitive Exams after 12th Science

## Access NCERT Solutions for Maths Class 11 Chapter 12 - Limits and Derivatives

### Exercise 12.1

1. Evaluate the Given limit.$\underset{x\to 3}{\mathop{\lim }}\,x+3$

Ans. Given,

$\underset{x\to 3}{\mathop{\lim }}\,x+3$

=$3+3$

=$6$

2. Evaluate the Given limit.$\underset{x\to \pi }{\mathop{\lim }}\,\left( x-\frac{22}{7} \right)$

Ans. Given,

$\underset{x\to \pi }{\mathop{\lim }}\,\left( x-\frac{22}{7} \right)$

=$\left( \pi -\frac{22}{7} \right)$

3. Evaluate the Given limit. $\underset{r\to 1}{\mathop{\lim }}\,\pi {{r}^{2}}$

Ans. Given,

$\underset{r\to 1}{\mathop{\lim }}\,\pi {{r}^{2}}$

=$\pi ({{1}^{2}})$

=$\pi$

4. Evaluate the Given limit.$\underset{x\to 1}{\mathop{\lim }}\,\frac{4x+3}{x-2}$

Ans. Given,

$\underset{x\to 1}{\mathop{\lim }}\,\frac{4x+3}{x-2}$

=$\frac{4(4)+3}{4-2}$

=$\frac{16+3}{2}$

=$\frac{19}{2}$

5. Evaluate the Given limit. $\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{10}}+{{x}^{5}}+1}{x-1}$

Ans. Given,

$\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{10}}+{{x}^{5}}+1}{x-1}$

=$\frac{{{(-1)}^{10}}+{{(-1)}^{5}}+1}{-1-1}$

=$\frac{1-1+1}{-2}$

=$-\frac{1}{2}$

6. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}$

Ans. Given,

$\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}$

Put $x+1=y$ So,

$y\to 1$ as $x\to 0$

Than, $\underset{x\to 0}{\mathop{\lim }}\,\frac{(x+1)_{{}}^{5}-1}{x}$

=$\underset{x\to 1}{\mathop{\lim }}\,\frac{(y)_{{}}^{5}-1}{y-1}$

Using $\left[ \underset{x\to a}{\mathop{\lim }}\,\frac{x_{{}}^{n}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]$

=${{5.1}^{5-1}}$

=$5$

7. Evaluate the Given limit. $\underset{x\to 2}{\mathop{\lim }}\,\frac{3{{x}^{2}}-x-10}{{{x}^{2}}-4}$

Ans. At $x=2$

The value of the given rational function takes the form $\frac{0}{0}$

=$\underset{x\to 2}{\mathop{\lim }}\,\frac{3{{x}^{2}}-x-10}{{{x}^{2}}-4}$

=$\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)(3x+5)}{(x-2)(x+2)}$

=$\underset{x\to 2}{\mathop{\lim }}\,\frac{3x+5}{x+2}$

=$\frac{3(2)+5}{2+2}$

=$\frac{11}{4}$

8. Evaluate the Given limit. $\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{4}}-81}{2{{x}^{2}}-5x-3}$

Ans. At $x=2$

The value of the given rational function takes the form $\frac{0}{0}$

=$\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{4}}-81}{2{{x}^{2}}-5x-3}$

=$\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)(x+3)({{x}^{2}}+9)}{(x-3)(2x+1)}$

=$\underset{x\to 3}{\mathop{\lim }}\,\frac{(x+3)({{x}^{2}}+9)}{(2x+1)}$

=$\frac{(3+3)({{3}^{2}}+9)}{(2(3)+1)}$

=$\frac{6\times 18}{7}$

=$\frac{108}{7}$

9. Evaluate the Given limit.$\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+b}{cx+1}$

Ans. Given,

$\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+b}{cx+1}$

=$\frac{a(0)+b}{c(0)+1}$

=$b$

10. Evaluate the Given limit.$\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{{{z}^{\frac{1}{6}}}-1}$

Ans. At $z=1$

The value of the given rational function takes the form $\frac{0}{0}$

Put ${{z}^{\frac{1}{6}}}=x$ So,

$z\to 1$ as $x\to 1$

Than,$\underset{z\to 1}{\mathop{\lim }}\,\frac{{{z}^{\frac{1}{3}}}-1}{{{z}^{\frac{1}{6}}}-1}$

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

Using $\left[ \underset{x\to a}{\mathop{\lim }}\,\frac{x_{{}}^{n}-{{a}^{n}}}{x-a}=n{{a}^{n-1}} \right]$

=${{2.1}^{2-1}}$

=$2$

11. Evaluate the Given limit. $\underset{x\to 1}{\mathop{\lim }}\,\frac{a{{x}^{2}}+bx+c}{c{{x}^{2}}+bx+a}$,$a+b+c\ne 0$

Ans. Given,

$\underset{x\to 1}{\mathop{\lim }}\,\frac{a{{x}^{2}}+bx+c}{c{{x}^{2}}+bx+a}$

=$\frac{a{{(1)}^{2}}+b(1)+c}{c{{(1)}^{2}}+b(1)+a}$

=$\frac{a+b+c}{c+b+a}$

=$1$                    $(a+b+c\ne 0)$

12. Evaluate the Given limit. $\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}$

Ans. Given,

$\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}$

At $x=-2$

The value of the given rational function takes the form $\frac{0}{0}$

$\underset{x\to -2}{\mathop{\lim }}\,\frac{\frac{1}{x}+\frac{1}{2}}{x+2}$=$\underset{x\to -2}{\mathop{\lim }}\,\frac{\left( \frac{2+x}{2x} \right)}{x+2}$

=$\underset{x\to -2}{\mathop{\lim }}\,\frac{1}{2x}$

=$\frac{1}{2(-2)}$

=$-\frac{1}{4}$

13. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}$

Ans. Given,

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}$

At $x=0$

The value of the given rational function takes the form $\frac{0}{0}$

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{bx}\times \frac{ax}{ax}$

=$\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{bx} \right)\times \frac{a}{b}$

=$\frac{a}{b}\underset{ax\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{bx} \right)$

$x\to 0\Rightarrow ax\to 0$

=$\frac{a}{b}\times 1$         $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right) \right]$

=$\frac{a}{b}$

14. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx},a,b\ne 0$

Ans.Given,

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx},a,b\ne 0$

At $x=0$

The value of the given rational function takes the form $\frac{0}{0}$

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax}{\sin bx}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{\sin ax}{ax} \right)\times ax}{\left( \frac{\sin ax}{ax} \right)\times bx}$

=$\frac{a}{b}\times \frac{\underset{ax\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{ax} \right)}{\underset{bx\to 0}{\mathop{\lim }}\,\left( \frac{\sin ax}{ax} \right)}$         \left[ \begin{align} & x\to 0\Rightarrow ax\to 0 \\ & x\to 0\Rightarrow bx\to 0 \\ \end{align} \right]

=$\frac{a}{b}\times \frac{1}{1}$                   $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

=$\frac{a}{b}$

15. Evaluate the Given limit. $\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}$

Ans. Given,

$\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}$

$\left[ x\to \pi \Rightarrow (\pi -x)\to 0 \right]$

$\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sin (\pi -x)}{\pi (\pi -x)}=\frac{1}{\pi }\underset{(\pi -x)\to 0}{\mathop{\lim }}\,\frac{\sin (\pi -x)}{(\pi -x)}$

$=\frac{1}{\pi }\times 1$                      $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

$=\frac{1}{\pi }$

16. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,\frac{cosx}{\pi -x}$

Ans. Given,

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\operatorname{cosx}}{\pi -x}$

=$\frac{\cos 0}{\pi -0}$

=$\frac{1}{\pi }$

17. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}$

Ans. Given,

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}$

At $x=0$

The value of the given rational function takes the form $\frac{0}{0}$

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos 2x-1}{\cos x-1}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-2{{\sin }^{2}}x-1}{1-2{{\sin }^{2}}\frac{x}{2}-1}$

=$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x}{{{\sin }^{2}}\frac{x}{2}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)\times {{x}^{2}}}{\left( \frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)\times \frac{{{x}^{2}}}{4}}$

=$4\frac{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)}$

=$4\frac{{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x}{{{x}^{2}}} \right)}^{2}}^{{}}}{{{\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}\frac{x}{2}}{{{\left( \frac{x}{2} \right)}^{2}}} \right)}^{2}}}$

$\left[ x\to 0\Rightarrow \frac{x}{2}\to 0 \right]$

=$4\frac{{{1}^{2}}^{{}}}{{{1}^{2}}}$         $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin y}{y} \right)=1 \right]$

=$4$

18. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}$

Ans.Given,

$\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}$

At $x=0$

The value of the given rational function takes the form $\frac{0}{0}$

$\underset{x\to 0}{\mathop{\lim }}\,\frac{ax+x\cos x}{b\sin x}=\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\frac{x(a+\cos x)}{\sin x}$

$\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{x}{\sin x} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right)$

=$\frac{1}{b}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin x}{x} \right)} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right)$

=$\frac{1}{b}\times \left( a+\cos 0 \right)$               $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

=$\frac{a+1}{b}$

19. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,x\sec x$

Ans. Given,

$\underset{x\to 0}{\mathop{\lim }}\,x\sec x$

=$\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\cos x}$

=$\frac{0}{\cos 0}$

=$\frac{0}{1}$

=$0$

20. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax+bx}{ax+\sin bx}$$a,b,a+b\ne 0$

Ans. At $x=0$

The value of the given rational function takes the form $\frac{0}{0}$

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ax+bx}{ax+\sin bx}$

=$\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{\sin ax}{ax} \right)ax+bx}{ax+bx\left( \frac{\sin bx}{bx} \right)}$

=$\frac{\left( \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin ax}{ax} \right)\times \underset{x\to 0}{\mathop{\lim }}\,(ax)+\underset{x\to 0}{\mathop{\lim }}\,(bx)}{\underset{x\to 0}{\mathop{\lim }}\,ax+\underset{x\to 0}{\mathop{\lim }}\,bx\left( \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin bx}{bx} \right)}$

$\left[ x\to \pi \Rightarrow ax\to 0 \right]$ and $\left[ bx\to 0 \right]$

=$\frac{\underset{x\to 0}{\mathop{\lim }}\,(ax)+\underset{x\to 0}{\mathop{\lim }}\,(bx)}{\underset{x\to 0}{\mathop{\lim }}\,ax+\underset{x\to 0}{\mathop{\lim }}\,bx}$         $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

=$\frac{\underset{x\to 0}{\mathop{\lim }}\,(ax+bx)}{\underset{x\to 0}{\mathop{\lim }}\,(ax+bx)}$

=$\underset{x\to 0}{\mathop{\lim }}\,(1)$

=$1$

21. Evaluate the Given limit. $\underset{x\to 0}{\mathop{\lim }}\,(\operatorname{cosecx}-cotx)$

Ans. At $x=0$

The value of the given rational function takes the form $\infty \to \infty$

$\underset{x\to 0}{\mathop{\lim }}\,(\operatorname{cosecx}-cotx)$

=$\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{\sin x}-\frac{\cos x}{\sin x} \right)$

=$\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos x}{\sin x} \right)$

=$\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{1-\cos x}{x} \right)}{\left( \frac{\sin x}{x} \right)}$

=$\frac{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos x}{x} \right)}{\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\sin x}{x} \right)}$

$\left[ \underset{y\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{x}=0 \right]$ and $\left[ \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

=$\frac{0}{1}$

=$0$

22. Evaluate the Given limit. $\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}$

Ans. Given,

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}$

At $x=\frac{\pi }{2}$

The value of the given rational function takes the form $\frac{0}{0}$

Put $x-\frac{\pi }{2}=y$

So,$\left[ x\to \frac{\pi }{2},y\to 0 \right]$

$\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 2x}{x-\frac{\pi }{2}}=\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan 2\left( y+\frac{\pi }{2} \right)}{y}$

=$\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan (\pi +2y)}{y}$

=$\underset{y\to 0}{\mathop{\lim }}\,\frac{\tan 2y}{y}$    $\left[ \tan (\pi +2y)=\tan 2y \right]$

=$\underset{y\to 0}{\mathop{\lim }}\,\frac{\sin 2y}{y\cos 2y}$

=$\underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\sin 2y}{2y}\times \frac{2}{\cos 2y} \right)$

=$\underset{y\to 0}{\mathop{\lim }}\,\left( \frac{\sin 2y}{2y} \right)\times \underset{y\to 0}{\mathop{\lim }}\,\left( \frac{2}{\cos 2y} \right)$  $\left[ y\to 0\Rightarrow 2y\to 0 \right]$

=$1\times \frac{2}{\cos 0}$    $\left[ \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\operatorname{sinx}}{x} \right)=1 \right]$

=$1\times \frac{2}{1}$

=$2$

23. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$and $\underset{x\to 1}{\mathop{\lim }}\,f(x)$, where f(x)=\left\{ \begin{align} & 2x+3 \\ & 3(x+1) \\ \end{align} \right.  \begin{align} & x\le 0 \\ & x>0 \\ \end{align}

Ans. Given.

f(x)=\left\{ \begin{align} & 2x+3 \\ & 3(x+1) \\ \end{align} \right.    \begin{align} & x\le 0 \\ & x>0 \\ \end{align}

$\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\left[ 2x+3 \right]$

=$2(0)+3=3$

$\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\left[ 3x+1 \right]$

=$3(0+1)=3$

Therefore,

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,f(x)=3$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,(x+1)=3(1+1)=6$

Therefore,

$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(x)=6$

24. Find $\underset{x\to 1}{\mathop{\lim }}\,f(x)$, where f(x)=\left\{ \begin{align} & {{x}^{2}}-1 \\ & -x-1 \\ \end{align} \right.   \begin{align} & x\le 1 \\ & x>1 \\ \end{align}

Ans. Given,