Question

If $f\left(x\right)=x\sin \left(1/x\right),x\ne 0,$ then $\underset{x\to 0}{\lim }f\left(x\right)=$
(a) 1
(b) 0
(c) −1
(d) does not exist

Answer 1

(b) 0
$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }x\sin \frac{1}{x},x\ne 0 \\ \mathrm{LHL}\mathrm{at}x=0: \\ \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right) \\ \Rightarrow \underset{h\to 0}{\lim }-h\sin \left(\frac{1}{-h}\right) \\ =0 \\ \mathrm{RHL}\mathrm{at}x=0: \\ \underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right) \\ \Rightarrow \underset{h\to 0}{\lim }h\sin \left(\frac{1}{h}\right) \\ =0 \\ \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right) \\ \mathrm{Hence},\underset{x\to 0}{\lim }f\left(x\right)\mathrm{exists}\mathrm{and}\mathrm{is}\mathrm{equal}\mathrm{to}0. \end{gathered}$$