Question

If $ f(x) = \begin{cases} x\,\sin\,\frac{1}{x}, & x \ne 0 \\[2ex] k, & x = 0 \end{cases} $ is continuous at $ x = 0 $ , then the value of $ k $ will be

• A. 1
• B. -1
• C. 0
• D. None of these

Answer 1

Answer: (C)

Given that $f(x)=\begin{cases}x \sin \frac{1}{x}, & x \neq 0 \\ k & , x=0\end{cases}$
$ LHL =\lim\limits _{x \rightarrow 0^{-}} f(x)$
$ =\lim\limits _{x \rightarrow 0^{-}} x \sin \frac{1}{x} $
$=\lim\limits _{h \rightarrow 0}(0-h) \sin \frac{1}{(0-h)} $
$=\lim\limits _{h \rightarrow 0} \,h \,\sin \frac{1}{h}=0 $
Since, $f(x)$ is continuous at $x=0$
$\therefore LHL =f(0)$
$\Rightarrow 0=k$