Задача №1143

\[ \lim_{x\to{-1}}\frac{\sqrt{\pi}-\sqrt{\arccos{x}}}{\sqrt{x+1}} =\left[\frac{0}{0}\right] =\lim_{t\to\pi}\frac{\sqrt{\pi}-\sqrt{t}}{\sqrt{1+\cos{t}}}=\\ =\lim_{t\to\pi}\frac{\left(\sqrt{\pi}-\sqrt{t}\right)\cdot\left(\sqrt{\pi}+\sqrt{t}\right)}{\sqrt{2\cos^2\frac{t}{2}}\cdot\left(\sqrt{\pi}+\sqrt{t}\right)} =\lim_{t\to\pi}\frac{\pi-t}{\sqrt{2}\cdot\cos\frac{t}{2}\cdot\left(\sqrt{\pi}+\sqrt{t}\right)} =\left[\begin{aligned}z=t-\pi;\\z\to{0}.\end{aligned}\right]=\\ =\lim_{z\to{0}}\frac{z}{\sqrt{2}\sin\frac{z}{2}\cdot\left(\sqrt{\pi}+\sqrt{z+\pi}\right)} =\lim_{z\to{0}}\frac{2}{\sqrt{2}\cdot\frac{\sin\frac{z}{2}}{\frac{z}{2}}\cdot\left(\sqrt{\pi}+\sqrt{z+\pi}\right)} =\frac{1}{\sqrt{2\pi}}. \]