Question

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

Answer 1

$\underset{x\to 0}{\overset{lim}{\to }}\frac{2\sin x}{x^2}\left[\frac{0}{0}form\right]$

Applying L'Hospital's Rule,

$\underset{x\to 0}{\overset{lim}{\to }}\frac{-2\cos x}{2x}=\underset{x\to 0}{\overset{lim}{\to }}\frac{-\cos x}{x}$

$=\frac{-1}{0}=$not defined

Ans=not defined.