Question

The value of $\underset{x\to 0}{lim}\frac{x^2\sin \frac{1}{x}}{\sin x}$ is?

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. None of the above

Answer 1

The correct option is B $0$

${lim}_{x\to 0}\frac{x^2\sin \frac{1}{x}}{\sin x}={lim}_{x\to 0}\frac{x}{\sin x}\times x\sin \frac{1}{x}$

$=1\times {lim}_{x\to 0}x\sin \frac{1}{x}$

We know that

$-1\le \sin \frac{1}{x}\le 1$

$-x\le x\sin \frac{1}{x}\le x$

${lim}_{x\to 0}(-x)\le {lim}_{x\to 0}x\sin \frac{1}{x}\le {lim}_{x\to 0}x$

$0\le {lim}_{x\to 0}x\sin \frac{1}{x}\le 0$

$\Rightarrow {lim}_{x\to 0}x\sin \frac{1}{x}=0$