**Question:**

If $f(x)=x^{2} \sin \frac{1}{x}$, where $x \neq 0$, then the value of the function $f$ at $x=0$, so that the function is continuous at $x=0$, is

(a) 0

(b) $-1$

(c) 1

(d) none

**Solution:**

The given function is $f(x)=x^{2} \sin \frac{1}{x}$, where $x \neq 0$.

Now, $f(x)$ is continuous at $x=0$.

$\therefore f(0)=\lim _{x \rightarrow 0} f(x)$

$\Rightarrow f(0)=\lim _{x \rightarrow 0} x^{2} \sin \frac{1}{x}$

$\Rightarrow f(0)=\lim _{x \rightarrow 0} x^{2} \times \lim _{x \rightarrow 0} \sin \frac{1}{x}$

$\Rightarrow f(0)=0 \times$ a finite value between $-1$ and 1 $\left(-1 \leq \sin \frac{1}{x} \leq 1\right)$

$\Rightarrow f(0)=0$

Thus, the value of the function $f$ at $x=0$ so that the function is continuous at $x=0$ is 0 .

Hence, the correct answer is option (a).