Question

If $f\left(x\right)=xsin\left(\frac{1}{x}\right),x\ne 0,$ then ${lim}_{x\to 0}f\left(x\right)=$

• A. 1
• B. 0
• C. -1
• D. does not exist

Answer 1

The correct option is

B

0

${lim}_{x\to 0}f\left(x\right)=limxsin\frac{1}{x},x\ne 0$
LHL at x = 0 :
$={lim}_{x\to 0}f\left(x\right)={lim}_{h\to 0}f(0-h)$
$\Rightarrow lim(-h)sin\left(\frac{1}{0-h}\right)$
$=0$ as multiplication of 0 and a finite quantity is always zero.
$RHLatx=0:$
$\Rightarrow {lim}_{h\to 0+}f\left(x\right)={lim}_{h\to 0}f(0+h)$
$\Rightarrow {lim}_{h\to 0}hsin\left(\frac{1}{h}\right)=0$ as multiplication of 0 and a finite quantity is always zero
$\Rightarrow {lim}_{h\to 0^{-}}f\left(x\right)$ =${lim}_{x\to 0^{+}}f\left(x\right)$
Therefore, ${lim}_{x\to 0}f\left(x\right)=0$