Chapter 13: Limits and Derivatives Exercise 13.1

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 $

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

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

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

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

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

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

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