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

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

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Access NCERT solutions for Class 11 Maths Chapter 13 –  Limits and Derivatives part-1

Access NCERT solutions for Class 11 Maths Chapter 13 – Limits and Derivatives

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 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.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 13 Exercise 13.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 13 Exercise 13.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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