Question

If $f\left(x\right)=\{\begin{array}{ccc}x\sin \left(1/x\right)& for& x\ne 0\\ 0& for& x=0\end{array}$ then

• A. $f$ is continuous but not differentiable
• B. $f$ is both continuous and differentiable
  
• C. $f$ is not continuous function
• D. $f$ is neither continuous nor differentiable

Answer 1

The correct option is A $f$ is continuous but not differentiable

$\begin{array}{c}f\left(x\right)=\left\{\frac{\sin \left(\frac{1}{x}\right)}{\left(1/x\right)}\right\}forx\ne 0\\ \Rightarrow f\left(o\right)=0\\ \Rightarrow f\left(o^{+}\right)f\left(o+h\right)=\sin \left(\frac{1}{\left(o+h\right)}\right)={lim}_{h\to 0}\frac{\sin \left(1/x\right)}{\left(1/x\right)}=1\\ \Rightarrow f\left(o^{-}\right)=f\left(o-h\right)={lim}_{h\to 0}\frac{\sin \frac{1}{\left(o+h\right)}}{\left(o+h\right)}=-\frac{\sin \frac{1}{h}}{-\frac{1}{h}}=1\\ \Rightarrow R{f}^{\prime }\left(o\right)={lim}_{h\to 0}\frac{f\left(o+h\right)-f\left(o\right)}{h}={lim}_{h\to 0}\left\{\frac{\sin \left(\frac{1}{o+h}-0\right)}{}\right\}={lim}_{h\to 0}\frac{h\sin 1/h-0}{h}\\ \Rightarrow {lim}_{h\to 0}\sin \left(\frac{1}{h}\right),doesnotexist\\ f\left(x\right)iscontinuousbutnotdifferentiatable\end{array}$