Question

The function $f\left(x\right)=\{\begin{array}{cc}x\sin \left(\ln x^2\right),& x\ne 0\\ 0,& x=0\end{array}$ is:

• A. non differentiable at $x=0$
• B. differentiable at $x=0$
• C. discontinuous at $x=0$
• D. none of the above

Answer 1

The correct option is A non differentiable at $x=0$

Given : $f\left(x\right)=\{\begin{array}{cc}x\sin \left(\ln x^2\right),& x\ne 0\\ 0,& x=0\end{array}$

For continuity,
$R.H.L=\underset{x\to 0^{+}}{lim}x\sin \left(\ln x^2\right)=0(\because \sin \theta \in [-1,1\left]\right)L.H.L.=\underset{x\to 0^{-}}{lim}x\sin \left(\ln x^2\right)=0\Rightarrow f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)=0$
Hence, $f\left(x\right)$ is continuous at $x=0$

For differentiability,
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{h\sin \left(\ln h^2\right)-0}{h}\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\sin \left(\ln h^2\right)\Rightarrow f^{\prime }\left(0^{+}\right)=\text{Not defined}$
Similarly, $L.H.D.$ is also not defined
Hence, $f\left(x\right)$ is not differentiable at $x=0$