Question

$\underset{x\to \mathrm{\pi }}{\lim }\frac{\sqrt{2+\cos x}-1}{{\left(\mathrm{\pi }-x\right)}^2}$

Answer 1

$$\begin{gathered} \underset{x\to \mathrm{\pi }}{\lim }\frac{\sqrt{2+\cos x}-1}{{\left(\mathrm{\pi }-x\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2+\cos \left(\mathrm{\pi }-h\right)}-1}{{\left(\mathrm{\pi }-\left(\mathrm{\pi }-h\right)\right)}^2} \\ =\underset{h\to 0}{\lim }\left(\frac{\sqrt{2-\cos h}-1}{h^2}\right)\times \frac{\sqrt{2-\cos h}+1}{\sqrt{2-\cos h}+1} \\ =\underset{h\to 0}{\lim }\frac{2-\cos h-1}{h^2\left(\sqrt{2-\cos h}+1\right)} \\ =\underset{h\to 0}{\lim }\frac{1-\cos h}{h^2\left(\sqrt{2-\cos h}+1\right)} \\ =\underset{h\to 0}{\lim }\frac{2{\sin }^2{\displaystyle \frac{h}{2}}}{4\times {\displaystyle \frac{h^2}{4}\times }\left[\sqrt{2-\cos h}+1\right]} \\ =\frac{1\times 1^2}{2\left(\sqrt{2-1}+1\right)} \\ =\frac{1}{4} \end{gathered}$$