Question

$\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)$ is equal to

• A. \N
• B. 1
• C. $\frac{1}{2}$
• D. does not exist

Answer 1

The correct option is A \N

$\mathrm{We}\mathrm{know}\mathrm{that}\underset{x\to 0}{\lim }x=0\mathrm{and}-1\le \sin \left(\frac{1}{x}\right)\le 1\mathrm{Sandwitch}\mathrm{Theorem}\mathrm{states}\mathrm{that}\mathrm{if}g\left(x\right),f\left(x\right)\mathrm{and}h\left(x\right)\mathrm{are}\mathrm{real}\mathrm{functions}\mathrm{such}\mathrm{that},g\left(x\right)\le f\left(x\right)\le h\left(x\right)\mathrm{then}\underset{x\to a}{\lim }g\left(x\right)\le \underset{x\to a}{\lim }f\left(x\right)\le \underset{x\to a}{\lim }h\left(x\right)\mathrm{Therefore},\underset{x\to 0}{\lim }-x\le \underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)\le \underset{x\to 0}{\lim }x\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)=0$