Since you already received good answers for the limit iteself.
You can go beyond the limit if you compose Taylor series (that is to say working one piece at the time).
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$
$$\sqrt{\cos(x)}=1-\frac{x^2}{4}-\frac{x^4}{96}+O\left(x^6\right)$$
$$\pi \sqrt{\cos(x)}=\pi -\frac{\pi x^2}{4}-\frac{\pi x^4}{96}+O\left(x^6\right)$$
$$\sin \left(\pi \sqrt{\cos (x)}\right)= \sin\left(\frac{\pi x^2}{4}+\frac{\pi x^4}{96}+O\left(x^6\right) \right)=\frac{\pi x^2}{4}+\frac{\pi x^4}{96}+O\left(x^6\right)$$
$$\frac{\sin(\pi\sqrt{\cos (x)})} x=\frac{\pi x}{4}+\frac{\pi x^3}{96}+O\left(x^5\right)$$ which shows the limit and how it is approached.
Moreover, this gives a shorcut evaluation of a not so pleasant expression. Suppose $x=\frac \pi{12}$ (this is quite far away from $0$.
The exact expression would be
$$\frac{\sin\left( \pi\sqrt{\frac {\sqrt 6+\sqrt 2} 4}\right)}{\frac \pi {12}} \approx 0.20612 $$ while the above truncated series gives
$$\frac{\pi ^2 \left(3456+\pi ^2\right)}{165888}\approx 0.20620$$