Question

If $f\left(x\right)=x^2\sin \frac{1}{x},$ where x ≠ 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

Answer 1

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

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

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

$\Rightarrow f\left(0\right)=\underset{x\to 0}{\lim }{x}^2\sin \frac{1}{x}$

$\Rightarrow f\left(0\right)=\underset{x\to 0}{\lim }{x}^2\times \underset{x\to 0}{\lim }\sin \frac{1}{x}$

⇒ f(0) = 0 × a finite value between − 1 and 1 $\left(-1\le \sin \frac{1}{x}\le 1\right)$

⇒ 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).