Question

Let $f(x)=\begin{cases}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{cases}$ Then at $x=0$

• A. $f$ is continuous but not differentiable
• B. $f$ is continuous but $f^{\prime}$ is not continuous
• C. $f^{\prime}$ is continuous but not differentiable
• D. $f$ and $f^{\prime}$ both are continuous

Answer 1

Answer: (B)

Continuity of $ f(x): f\left(0^{+}\right)=h^2 \cdot \sin \frac{1}{h}=0$
$ f\left(0^{-}\right)=(-h)^2 \cdot \sin \left(\frac{-1}{h}\right)=0 $
$ f(0)=0 $
$ f(x) $ is continuous
$ f^{\prime}\left(0^{+}\right)=\displaystyle\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\frac{h^2 \cdot \sin \left(\frac{1}{h}\right)-0}{h}=0 $
$f^{\prime}\left(0^{-}\right)=\displaystyle\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\frac{h^2 \cdot \sin \left(\frac{1}{-h}\right)-0}{-h}=0$
$f(x)$ is differentiable.
$f^{\prime}(x)=2 x \cdot \sin \left(\frac{1}{x}\right)+x^2 \cdot \cos \left(\frac{1}{x}\right) \cdot \frac{-1}{x^2}$
$f^{\prime}(x)=\begin{cases}2 x \cdot \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{cases}$
$\Rightarrow f^{\prime}(x)$ is not continuous (as $\cos \left(\frac{1}{x}\right)$ is highly oscillating at $x=0$ )