Question

If $f\left(x\right)=\{\begin{array}{ccc}x\sin \left(1/x\right)& for& x\ne 0\\ 0& for& x=0\end{array}$ then

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

Answer 1

The correct option is A $f$ is continuous function

Now

$\underset{x\to 0}{lim}f\left(x\right)=0$.

Since,

$\left|x\sin \frac{1}{x}\right|\le |x|,\forall x\in R$ [ since $\left|\sin \frac{1}{x}\right|\le 1$].

So $\underset{x\to 0}{lim}f\left(x\right)=0=f\left(0\right)$. [ By Sandwich theorem]

Hence the function is continuous at $x=0$.

Again,

$f^{\prime }\left(0\right)$

$=\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$

$=\underset{x\to 0}{lim}\sin \frac{1}{x}$.

This limit doesn't exist. By sequential criterion it can be shown that

$\underset{x\to 0}{lim}\sin \frac{1}{x}$ doesn't exist.