Evaluating $\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x$ I don't know how can this be solved:
$$
\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x
$$
I've tried to multiply by $\pi\sqrt{\cos x}$:
$$
\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})}{x\pi\sqrt{\cos x}}
$$
but we have here a hard problem $\lim_{x\to0}(1/x)$.
You think about the limit on the right and on the left but the exercise supposed to be solved by the limit in $0$ and not on the right or the left of it.

We know that:
$\sin(\alpha)=\sin(\pi-\alpha)$ and
$$\lim_{x\to 0}\frac{\sin(x)}x=1\quad\&\quad\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$$
$$\begin{aligned}\lim_{x\to 0}\frac{\sin(\pi\sqrt{\cos(x)})}x&=\lim_{x\to 0}\frac{\sin(\pi(1-\sqrt{\cos(x)}))}{\pi(1-\sqrt{\cos(x)})}\frac{(1-\sqrt{\cos(x)})}{x^2}\frac{1+\sqrt{\cos(x)}}{1+\sqrt{\cos(x)}}\pi x\\&=\lim_{x\to 0}\frac{\sin(\pi(1-\cos(x)))}{\pi(1-\sqrt{\cos(x)})}\frac{1-\cos(x)}{x^2}\frac{\pi x}{1+\sqrt{\cos(x)}}=0\end{aligned}$$
