Answer:
Option A
Explanation:
Given
$f(x)=\begin{cases}x \sin\frac{1}{x} ,&for x \neq 0\\k, & for x=0 \end{cases}$
iSince , f(x) is continuous at x=0
, $\therefore$ LHL=RHL=f(0)
.......(i)
Now, LHL$=\lim_{x \rightarrow 0-}f(x)=\lim_{h \rightarrow 0}f(0-h)$
$=\lim_{h \rightarrow 0}(0-h)\sin \frac{1}{(0-h)}=\lim_{h \rightarrow 0}-h\sin\left(-\frac{1}{h}\right)$
$=\lim_{h \rightarrow 0}h \sin \frac{1}{h}=0\times$ finite value =0
$\left[\because \lim_{h \rightarrow 0}\sin \frac{1}{h}=finite value\right]$
and f(0)=k
$\therefore$ From Eq.(i) , LHL= RHL
$\Rightarrow$ 0=k
$\Rightarrow$ k=0
