Answer:
Option C
Explanation:
For function to be continuous :
f(0+h) = f(0-h) = f(0)
f(0 + h) = $\lim_{h \rightarrow 0}h\sin1/h$ = 0 x (a finite quantity) = 0
f (0-h) = $\lim_{h \rightarrow 0}-h\sin1/-h$ = 0 x (a finite quantity) = 0
$\lim_{x \rightarrow 0}x\sin1/x$ = 0 x (a finite quantity) = 0
function is continuous at x = 0
For function to be differentiable :
f' (0 + h) = f'(0 - h)
$\frac{f(0+h)-f(0)}{h}$ = $\lim_{h \rightarrow 0}\frac{h \sin1/h - 0}{h}$
= $\lim_{h \rightarrow 0}\sin\left(\frac{ 1}{h}\right)$
which does not exist
