Chapter 12: Limits and Derivatives

Ans. Given,

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

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

=\[\frac{a+b+c}{c+b+a}\]

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

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 \[\frac{0}{0}\] 

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

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

=\[\frac{1}{2(-2)}\]

=\[-\frac{1}{4}\]

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 \[\frac{0}{0}\] 

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

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

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

$x\to 0\Rightarrow ax\to 0$ 

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

=\[\frac{a}{b}\]

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 \[\frac{0}{0}\] 

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

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

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

=\[\frac{a}{b}\]

Ans. Given,

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

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

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

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

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

Ans. Given,

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

=\[\frac{\cos 0}{\pi -0}\]

=\[\frac{1}{\pi }\]

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 \[\frac{0}{0}\] 

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

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

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

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

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

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

=\[4\]   

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 \[\frac{0}{0}\] 

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

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

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

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

  =\[\frac{a+1}{b}\]         

Ans. Given,

\[\underset{x\to 0}{\mathop{\lim }}\,x\sec x\]

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

  =\[\frac{0}{\cos 0}\]

=\[\frac{0}{1}\]

=\[0\]

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 \[\left[ bx\to 0 \right]\]

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

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

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

  =\[1\]