Question

If $f\left(x\right)=\{\begin{array}{cc}{x}^2\sin \frac{1}{x}& \text{for}x\ne 0\\ 0& \text{for}x=0\end{array}$ then

• A. f and $f^{\prime }$ are continuous at x=0
• B. f is derivable at x=0
• C. f is derivable at x=0 and $f^{\prime }$ is not continuous at x=0
• D. f is derivable at x=0 and $f^{\prime }$ is continuous at x=0.

Answer 1

Answer: (B), (C)

f is derivable at x=0
f is derivable at x=0 and $f^{\prime }$ is not continuous at x=0

For $x\ne 0$, we have
$f^{\prime }\left(x\right)=2x\sin \frac{1}{x}+x^2\cos \left(\frac{1}{x}\right)(-\frac{1}{x^2})$
$=2x\sin \frac{1}{x}-\cos \frac{1}{x}$
and for $x\ne 0$, we have
$\frac{f\left(x\right)-f\left(0\right)}{x-0}=\frac{x^2\sin \left(1/x\right)}{x}=x\sin \frac{1}{x}\Rightarrow f^{\prime }\left(0\right)=0$
Thus f is derivable at x=0. Also,
$\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0}{lim}(2x\sin \frac{1}{x}-\cos \frac{1}{x})$
Let us write
$\cos \frac{1}{x}=2x\sin \frac{1}{x}-(2x\sin \frac{1}{x}-\cos \frac{1}{x})$
Now ${lim}_{x\to 0}\left(2x\sin \left(1/x\right)\right)=0$, so that if ${lim}_{x\to 0}{f}^{\prime }\left(x\right)$ were to exist, then ${lim}_{x\to 0}\cos \left(1/x\right)$ would also exist, which is not true. Hence $f^{\prime }$ is not continuous at $x=0$.