Evaluating $\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x$

We know that: $\sin(\alpha)=\sin(\pi-\alpha)$ and

$$\lim_{x\to 0}\frac{\sin(x)}x=1\quad\&\quad\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$$

$$\begin{aligned}\lim_{x\to 0}\frac{\sin(\pi\sqrt{\cos(x)})}x&=\lim_{x\to 0}\frac{\sin(\pi(1-\sqrt{\cos(x)}))}{\pi(1-\sqrt{\cos(x)})}\frac{(1-\sqrt{\cos(x)})}{x^2}\frac{1+\sqrt{\cos(x)}}{1+\sqrt{\cos(x)}}\pi x\\&=\lim_{x\to 0}\frac{\sin(\pi(1-\cos(x)))}{\pi(1-\sqrt{\cos(x)})}\frac{1-\cos(x)}{x^2}\frac{\pi x}{1+\sqrt{\cos(x)}}=0\end{aligned}$$

Note that for $x\to 0$, $\dfrac {\sin(\pi\sqrt{\cos x})}{\pi\sqrt{\cos x}} \to \dfrac {\sin \pi}{\pi}=0$. The limit for $\dfrac {\sin x}x$ is $1$ for $x \to 0$ only.

Hence we transform the argument of the sine so it tends to $0$, by seeing

$$\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x = \lim_{x\to0} \frac{\sin(\pi - \pi\sqrt{\cos x})} x = \lim_{x\to0} \frac{\sin(\pi - \pi\sqrt{\cos x})} {\pi-\pi\sqrt{\cos x}}\cdot \frac {\pi-\pi\sqrt{\cos x}}x$$

Now the first part of the product has limit $1$. Do you know how to calculate the limit of the second part?

Since $\cos x=1-x^2/2+o(x^2)$, we also have $$ \sqrt{\cos x}=1-\frac{x^2}{4}+o(x^2) $$ and so your limit can be rewritten as $$ \lim_{x\to0}\frac{\sin(\pi-\pi x^2/4+o(x^2))}{x}=\lim_{x\to0}\frac{\sin(\pi x^2/4+o(x^2))}{x}=\lim_{x\to0}\frac{\pi x^2/4+o(x^2)}{x}=0 $$ With $x^2$ at the denominator you'd get $\pi/4$.

A different way to solve the problem is to notice that the given limit is the derivative of $f(x)=\sin(\pi\sqrt{\cos x})$ at $0$; since $$ f'(x)=\cos(\pi\sqrt{\cos x})\frac{-\pi\sin x}{2\sqrt{\cos x}} $$ we get $f'(x)=0$