Question

If $f\left(x\right)=\left\{\begin{array}{ll}x\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array},\right.$ then $\underset{x\to 0}{\lim }f\left(x\right)$ equals

(a) 1
(b) 0
(c) −1
(d) none of these

Answer 1

(b) 0

$f\left(x\right)=\left\{\begin{array}{l}x\sin \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}\right.$

LHL:

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right) \\ =\underset{x\to 0^{-}}{\lim }\left[x\sin \left(\frac{1}{x}\right)\right] \end{gathered}$$

Let x = 0 – h, where h → 0.

$=\underset{h\to 0}{\lim }\left[\left(-h\right)\times \sin \left(-\frac{1}{h}\right)\right]$

= 0 × The oscillating number between –1 and 1
= 0
RHL
$\underset{x\to 0^{+}}{\lim }f\left(x\right)$

Let x = 0 + h, where h → 0.

$=\underset{h\to 0}{\lim }\left[h\times \sin \left(\frac{1}{h}\right)\right]$

= 0 × The oscillating number between –1 and 1
= 0
LHL = RHL = 0
$\therefore \underset{x\to 0}{\lim }f\left(x\right)=0$