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

The Correct Option is 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
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{f(0+h)-f(0)}{h}=\frac{h^2\cdot \sin \left(\frac{1}{h}\right)-0}{h}=0$
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{f(0-h)-f(0)}{-h}=\frac{h^2\cdot \sin \left(\frac{1}{-h}\right)-0}{-h}=0$
$f(x)$ is differentiable.
$f^{\prime }(x)=2x\cdot \sin \left(\frac{1}{x}\right)+x^2\cdot \cos \left(\frac{1}{x}\right)\cdot \frac{-1}{x^2}$
$f^{\prime }(x)=\{\begin{array}{ll}2x\cdot \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}$
$\Rightarrow f^{\prime }(x)$ is not continuous (as $\cos \left(\frac{1}{x}\right)$ is highly oscillating at $x=0$ )