Q. Define $f$ on $R$ into itself by $f(x) = \left\{ \begin{array}{r} x\sin{\frac{1}{x},\ \ when\ x \neq 0} \\ 0,\ \ \ when\ x = 0 \end{array} \right.\$, then
A) $f$ is continuous at 0 but not differentiable at 0
B) $f$ is both continuous and differentiable at 0
C) $f$ is differentiable but not continuous at 0
D) None of the above
Explanation
(a)
LHL$= \lim_{x \rightarrow 0^{-}}{f(x) =}\lim_{h \rightarrow 0} - h\sin\left( \frac{1}{- h} \right) = 0$
RHL$= \lim_{x \rightarrow 0^{+}}{f(x) =}\lim_{h \rightarrow 0}h\sin\left( \frac{1}{h} \right) = 0$
$\therefore$ LHL=RHL$= f(0)$, it is continuous
LHD$= \lim_{x \rightarrow 0^{-}}{f(x) =}\lim_{h \rightarrow 0}\left\lbrack \frac{f(0 - h) - f(0)}{- h} \right\rbrack$
$= \lim_{h \rightarrow 0}\left\lbrack \frac{- h\sin{\frac{1}{h} - 0}}{- h} \right\rbrack$=does not exist
$\Rightarrow \ f(x)$ is not differentiable at $x = 0$
$\therefore\ \ f(x)$ is continuous at $x = 0$ but not differentiable at $x = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.