Chapter 13: Limits and Derivatives Exercise 13.1

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

  =\[\frac{0}{1}\]        

  =\[0\]               

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

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

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

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

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

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

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

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

  =\[1\times \frac{2}{1}\]

  =\[2\]

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,

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

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.

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.

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.

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

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 $b=4$

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,

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

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 $.