Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^2\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}.$ Then

• A. $f^{\prime }$ does not exist at $x=0$
• B. $f^{\prime }$ exists and is continuous at $x=0$
• C. $f^{\prime }$ exists but not continuous at $x=0$
• D. $f^{\prime }$ does not exist at any point

Answer 1

The correct option is C $f^{\prime }$ exists but not continuous at $x=0$

Clearly, $f^{\prime }\left(x\right)$ exists for all $x\ne 0$

$f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{h^2\sin \frac{1}{h}}{h}$
$=\underset{h\to 0}{lim}h\sin \frac{1}{h}=0$
$\therefore f^{\prime }\left(0\right)$ exists and equals $0$

When $x\ne 0,$
$f^{\prime }\left(x\right)=-\cos \frac{1}{x}+2x\sin \frac{1}{x}$
$\underset{h\to 0}{lim}{f}^{\prime }\left(0^{-}\right)$ and $\underset{h\to 0}{lim}{f}^{\prime }\left(0^{+}\right)$ do not exist.
Hence, $f^{\prime }$ is not continuous at $x=0$