Задача №1143
Условие
Найти \(\lim_{x\to{-1}}\frac{\sqrt{\pi}-\sqrt{\arccos{x}}}{\sqrt{x+1}}\).
Решение
Полагая \(x=\cos{t}\), \(t\to\pi\), получим:
\[
\lim_{x\to{-1}}\frac{\sqrt{\pi}-\sqrt{\arccos{x}}}{\sqrt{x+1}}
=\left[\frac{0}{0}\right]
=\lim_{t\to\pi}\frac{\sqrt{\pi}-\sqrt{t}}{\sqrt{1+\cos{t}}}=\\
=\lim_{t\to\pi}\frac{\left(\sqrt{\pi}-\sqrt{t}\right)\cdot\left(\sqrt{\pi}+\sqrt{t}\right)}{\sqrt{2\cos^2\frac{t}{2}}\cdot\left(\sqrt{\pi}+\sqrt{t}\right)}
=\lim_{t\to\pi}\frac{\pi-t}{\sqrt{2}\cdot\cos\frac{t}{2}\cdot\left(\sqrt{\pi}+\sqrt{t}\right)}
=\left[\begin{aligned}z=t-\pi;\\z\to{0}.\end{aligned}\right]=\\
=\lim_{z\to{0}}\frac{z}{\sqrt{2}\sin\frac{z}{2}\cdot\left(\sqrt{\pi}+\sqrt{z+\pi}\right)}
=\lim_{z\to{0}}\frac{2}{\sqrt{2}\cdot\frac{\sin\frac{z}{2}}{\frac{z}{2}}\cdot\left(\sqrt{\pi}+\sqrt{z+\pi}\right)}
=\frac{1}{\sqrt{2\pi}}.
\]
Ответ:
\(\frac{1}{\sqrt{2\pi}}\)