Question

If $f\left(x\right)=\{\begin{array}{c}x\sin \left(1/x\right),ifx\ne 0\\ 0,ifx=0\end{array}$ then at $x=0$ the function $f$ is

• A. Continuous but not differentiable
• B. Differentiable but not continuous
• C. Continuous and differentiable
• D. Not continuous

Answer 1

The correct option is A Continuous but not differentiable

To determine whether $f\left(x\right)$ is is continuous or not at x=0,

We need to find $f\left(0^{+}\right),f\left(0^{-}\right)$

Now, $f\left(0^{+}\right)=x\sin \frac{1}{x}$

as $x\to 0^{+},\sin \left(\frac{1}{x}\right)$ oscillates from $-1$ to $+1$

Hence, $\underset{x\to 0^{+}}{lim}x\sin \left(\frac{1}{x}\right)=0$

And $f\left(0^{-}\right)=x\sin \frac{1}{x}$

As $x\to 0^{-},\sin \left(\frac{1}{x}\right)$ oscillates from $-1$ to $+1$

Hence, $\underset{x\to 0^{-}}{lim}x\sin \left(\frac{1}{x}\right)=0$

We get,
$f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)$

Thus, the function is continuous at $x=0$

$f\left(x\right)=x\sin \frac{1}{x}forx\ne 0$

$⟹f^{\prime }\left(x\right)=\sin \frac{1}{x}-\frac{1}{x}\cos \frac{1}{x}$

But as $x\to 0$ the derivative of $f\left(x\right)$ becomes undefined.

$\therefore$ We can say that $f\left(x\right)$ is continuous at $x=0$ but not differentiable.