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
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $0$ ##
Given, $f(x) = x^2 \sin \left( \frac{1}{x} \right)$, where $x \neq 0 \quad \left[ ∵-1 \leq \sin \frac{1}{x} \leq 1, \forall x \in R \right]$
Both $x^2$ and $\sin \frac{1}{x}$ are continuous functions, $\forall x \in R$
Hence, value of the function $f$ at $x = 0$, so that it is continuous at $x = 0$ is $0$.