$\underset{x\to 0}{lim} \frac{\cos (\sin x)-1}{x^2}=$

Solution:

We have,

$\begin{array}{l}\underset{x\to 0}{lim} \frac{\cos (\sin x)-1}{x^2}\\ =-\underset{x\to 0}{lim} \frac{1-\cos (\sin x)}{{\sin }^{2}x}\times \frac{{\sin }^{2}x}{x^2}\end{array}$

$=-\frac{1}{2}\times 1 \left[\because \underset{x\to 0}{lim} \frac{1-\cos x}{x^2}=\frac{1}{2}\right]$