Question

lf $f\left(x\right)=x.\sin \frac{1}{x}$ for $x\ne 0,f\left(0\right)=0$ then?

• A. $f$ is continuous at $x=0$
• B. $f$ is differentiable at $x=0$
• C. $f^{\prime }\left(0^{-}\right)$ exists but $f^{\prime }\left(0^{+}\right)$ does not exist
• D. $f$ is discontinuous at $x=0$

Answer 1

Answer: (A)

If $f\left(x\right)$ is continuous at $x=0$, then $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$
$LHL=\underset{h\to 0}{lim}\left[\right(0-h\left)\sin \left(\frac{1}{0-h}\right)\right]=\underset{h\to 0}{lim}h\sin \frac{1}{h}=0\times numberbet.0and1=0$
$RHL=\underset{h\to 0}{lim}h\sin \frac{1}{h}=0\times number=0$
Hence $LHL=RHL=f\left(0\right)$. The given function is continuous at $x=0$
.
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{h\sin \frac{1}{h}-f\left(0\right)}{h}=\underset{h\to 0}{lim}\sin \frac{1}{h}-0=\underset{h\to 0}{lim}\sin \frac{1}{h}$
Since this limit does not exist, the given function is not differentiable at $x=0$