Question

Prove that:

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

Answer 1

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

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

$={lim}_{x\to 0}\frac{sin\frac{1}{x}}{\frac{1}{x}}$ $[\therefore {lim}_{x\to 0}\frac{x}{sinx}=1]$

$={lim}_{x\to 0}xsin\frac{1}{x}=0\times$ [A no. between $(-1,1)]=0$