\begin{aligned} \lim_{x\to \pi} \ \frac{\sqrt{5+\cos x}-2}{(\pi-x)^2} &= \lim_{x\to \pi} \ \frac{\sqrt{5+\cos x}-2}{(\pi-x)^2} \times \frac{\sqrt{5+\cos x}+2}{\sqrt{5+\cos x}+2} \\[8pt] &= \lim_{x\to \pi} \ \frac{(5+\cos x)-4}{(\pi-x)^2 (\sqrt{5+\cos x}+2)} \\[8pt] &= \lim_{x\to \pi} \ \frac{1 + \cos x}{(\pi-x)^2 (\sqrt{5+\cos x}+2)} \times \frac{1 - \cos x}{1 - \cos x} \\[8pt] &= \lim_{x\to \pi} \ \frac{1 - \cos^2 x}{(\pi-x)^2 (\sqrt{5+\cos x}+2)(1 - \cos x)} \\[8pt] &= \lim_{x\to \pi} \ \frac{\sin^2 x}{(\pi-x)^2 (\sqrt{5+\cos x}+2)(1 - \cos x)} \\[8pt] &= \lim_{x\to \pi} \ \frac{\sin^2 (\pi-x)}{(\pi-x)^2 (\sqrt{5+\cos x}+2)(1 - \cos x)} \\[8pt] &= \lim_{x\to \pi} \ \frac{\sin^2 (\pi-x)}{(\pi-x)^2} \cdot \lim_{x\to \pi} \ \frac{1}{(\sqrt{5+\cos x}+2)(1 - \cos x)} \\[8pt] &= 1 \cdot \frac{1}{(\sqrt{5+\cos \pi}+2)(1 - \cos \pi)} = \frac{1}{(\sqrt{5+(-1)}+2)(1 - (-1))} \\[8pt] &= \frac{1}{(\sqrt{4}+2)(2)} = \frac{1}{8} \end{aligned}