NCERT Solutions for Class 11 Maths Chapter 13: Limits and Derivatives - Exercise 13.1

Exercise 13.1

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, 

\[f(x)=\left\{ \begin{align} & {{x}^{2}}-1 \\ & -x-1 \\ \end{align} \right.\]     $$\begin{align} & x\le 1 \\ & x>1 \\ \end{align}$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,({{x}^{2}}-1)={{1}^{2}}-1=1-1=0$

So, it is observed that 

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

Hence, $\underset{x\to {{1}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.

25. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & \frac{\left| x \right|}{x} \\ & 0 \\ \end{align} \right.\]    $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

Ans. Given

\[f(x)=\left\{ \begin{align} & \frac{\left| x \right|}{x} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{\left| x \right|}{x} \right]$

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{-x}{x} \right)\]    when x is negative, $\left| x \right|=-x$ 

=\[\underset{x\to 0}{\mathop{\lim }}\,(-1)\]

=\[-1\]

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

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

 when x is positive, $\left| x \right|=x$ 

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

=\[1\]

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

Hence, $\underset{x\to {{0}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.

26. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$, where \[f(x)=\left\{ \begin{align} & \frac{x}{\left| x \right|} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$

Ans. Given

\[f(x)=\left\{ \begin{align} & \frac{x}{\left| x \right|} \\ & 0 \\ \end{align} \right.\] $\begin{align} & x\ne 0 \\ & x=0 \\ \end{align}$ 

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{x}{\left| x \right|} \right]$

=\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{-x}{x} \right)\]    when x is negative, $\left| x \right|=-x$ 

=\[\underset{x\to 0}{\mathop{\lim }}\,(-1)\]

=\[-1\]

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

 when x is positive, $\left| x \right|=x$ 

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

=\[1\]

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

Hence, $\underset{x\to {{0}^{{}}}}{\mathop{\lim }}\,f(x)$ does not exist.

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

Ans. Given,

 $f(x)=\left| x \right|-5$

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,(\left| x \right|-5)$ 

= $\underset{x\to 5}{\mathop{\lim }}\,(x-5)$    when x is positive, $\left| x \right|=x$ 

= $5-5$ 

= $0$ 

$\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,(\left| x \right|-5)$ 

= $\underset{x\to 5}{\mathop{\lim }}\,(x-5)$ 

when x is positive, $\left| x \right|=x$ 

= $5-5$ 

= $0$ 

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f(x)$

Hence, $\underset{x\to 5}{\mathop{\lim }}\,f(x)=0$ 

28. Suppose $f(x)=\left\{ \begin{align} & a+bx \\ & 4 \\ & b-ax \\ \end{align} \right.$   $\begin{align} & x<0 \\ & x=1 \\ & x>1 \\ \end{align}$and if $\underset{x\to 1}{\mathop{\lim }}\,f(x)=f(1)$what are possible values of a and b?

Ans. The given function is

$f(x)=\left\{ \begin{align} & a+bx \\ & 4 \\ & b-ax \\ \end{align} \right.$ $\begin{align} & x<0 \\ & x=1 \\ & x>1 \\ \end{align}$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(ax+bx)=a+b$

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,f(b-ax)=b-a$

$f(1)=4$

Given

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

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

$a+b=4$and $b-a=4$

On solving,we get

$a=0$and $b=4$

29. Let a1,a2,……an be fixed real number and define a fuction

$f(x)=(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})$ 

What is $\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,f(x)$ ? For some a$\ne $ a1, a2,…..,an. Compute $\underset{x\to a}{\mathop{\lim }}\,f(x)$.

Ans. Given,

$f(x)=(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})$

$\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}_{1}}}{\mathop{\lim }}\,[(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})]$

=\[({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})=0\]

Therefore,

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

Now, $\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{x\to a}{\mathop{\lim }}\,[(x-{{a}_{1}})(x-{{a}_{2}})....(x-{{a}_{n}})]$

=\[({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})\]

Therefore,

$\underset{x\to a}{\mathop{\lim }}\,f(x)=({{a}_{1}}-{{a}_{1}})({{a}_{1}}-{{a}_{2}})......({{a}_{1}}-{{a}_{n}})$

30. If \[f(x)=\left\{ \begin{align} & \left| x \right|+1 \\ & 0 \\ & \left| x \right|-1 \\ \end{align} \right.\] $\begin{align} & x<0 \\ & x=0 \\ & x>1 \\ \end{align}$

For what value (s) of does $\underset{x\to a}{\mathop{\lim }}\,f(x)$  exists? 

Ans. Given,

 \[f(x)=\left\{ \begin{align} & \left| x \right|+1 \\ & 0 \\ & \left| x \right|-1 \\ \end{align} \right.\] $\begin{align} & x<0 \\ & x=0 \\ & x>1 \\ \end{align}$ 

When $a=0$ 

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

=\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -x+1 \right)\]     when x is negative, $\left| x \right|=-x$ 

=$0+1$

=$1$ 

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

=\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( x-1 \right)\]     when x is positive, $\left| x \right|=x$ 

=$0-1$

=$-1$ 

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

$\underset{x\to 0}{\mathop{\lim }}\,f(x)$ does not exist.

When\[\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,(\left| x \right|+1)\] $a<0$

\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(\left| x \right|+1)\]

=\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(-x+1)\]           $[x<a<0\Rightarrow \left| x \right|=-x]$

=\[-a+1\]

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

Thus, limit exists at $x=a$, where \[a<0\]

When \[a>0\]

$\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,(\left| x \right|+1)$

=$\underset{x\to a}{\mathop{\lim }}\,(-x-1)$                         $[0<x<a\Rightarrow \left| x \right|=x]$

=$-a-1$

$\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,(\left| x \right|+1)$

=$\underset{x\to a}{\mathop{\lim }}\,(-x-1)$                         $[0<x<a\Rightarrow \left| x \right|=x]$

=$a-1$

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

Thus, limit exists at $x=a$, where \[a>0\]

Thus $\underset{x\to a}{\mathop{\lim }}\,f(x)$ exists for all $a\ne 0$$\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi $.

31. If function f(x) satisfies, \[\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi \], evaluate $\underset{x\to 1}{\mathop{\lim }}\,f(x)$

Ans. Given, 

\[\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-2}{{{x}^{2}}-1}=\pi \]

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

= $\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=\pi \underset{x\to 1}{\mathop{\lim }}\,({{x}^{2}}-1)$

=$\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=\pi ({{1}^{2}}-1)$

=$\underset{x\to 1}{\mathop{\lim }}\,(f(x)-2)=0$

=$\underset{x\to 1}{\mathop{\lim }}\,f(x)-\underset{x\to 1}{\mathop{\lim }}\,2=0$

=$\underset{x\to 1}{\mathop{\lim }}\,f(x)-2=0$

$\underset{x\to 1}{\mathop{\lim }}\,f(x)=2$

32. If $f(x)=\left\{ \begin{align} & m{{x}^{2}}+n \\ & nx+m \\ & n{{x}^{3}}+m \\ \end{align} \right.$ $\begin{align} & x<0 \\ & 0\le x\le 1 \\ & x>1 \\ \end{align}$

For what integers m and n does $\underset{x\to 0}{\mathop{\lim }}\,f(x)$ and $\underset{x\to 1}{\mathop{\lim }}\,f(x)$ exists?

Ans. Given,  

$f(x)=\left\{ \begin{align} & m{{x}^{2}}+n \\ & nx+m \\ & n{{x}^{3}}+m \\ \end{align} \right.$ $\begin{align} & x<0 \\ & 0\le x\le 1 \\ & x>1 \\ \end{align}$\[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,(m{{x}^{2}}+n)\]

=$m{{(0)}^{2}}+n$

=$n$

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

=$n{{(0)}^{{}}}+m$

=$m$

Thus, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)$exists if m=n

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

=$n(1)+m$

=$m+n$

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

=$n{{(1)}^{3}}+m$

=$m+n$

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

Thus, $\underset{x\to 1}{\mathop{\lim }}\,f(x)$exists for any internal value of  m and n

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 13 All Exercises

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Exercise 13.1

Opting for the NCERT solutions for Ex 13.1 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.1 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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