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
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).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.