$\begin{aligned} & \text { Put } \cos ^{-1} x=y \text {. So if } x \rightarrow-1, y \rightarrow \pi \\ & \therefore \lim _{x \rightarrow-1} \frac{\sqrt{\pi}-\sqrt{\cos ^{-1} x}}{\sqrt{x+1}}=\lim _{y \rightarrow \pi} \frac{\sqrt{\pi}-\sqrt{y}}{\sqrt{1+\cos y}} \\ & =\lim _{y \rightarrow \pi} \frac{\sqrt{\pi}-\sqrt{y}}{\sqrt{2} \cos (y / 2)}=\lim _{y \rightarrow \pi} \frac{\sqrt{\pi}-\sqrt{y}}{\sqrt{2} \sin \left(\frac{\pi}{2}-\frac{y}{2}\right)} \frac{\left(\frac{\pi}{2}-\frac{y}{2}\right)}{\left(\frac{\pi}{2}-\frac{y}{2}\right)} \\ & =\lim _{y \rightarrow \pi} \frac{1}{\frac{\sqrt{2}}{2}(\sqrt{\pi}+\sqrt{y})} \cdot \frac{1}{\frac{\sin \left(\frac{\pi}{2}-\frac{y}{2}\right)}{\left(\frac{\pi}{2}-\frac{y}{2}\right)}}=\frac{1}{\sqrt{2 \pi}} .\end{aligned}$