Задача №1181
Условие
Найти предел \(\lim_{x\to\frac{\pi}{2}}\tg^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\).
Решение
\[
\lim_{x\to\frac{\pi}{2}}\tg^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)=\\
=\lim_{x\to\frac{\pi}{2}}\left(\frac{\sin^2{x}}{\cos^2{x}}\cdot\frac{\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\cdot\left(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\right)}{\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}}\right)=\\
=\lim_{x\to\frac{\pi}{2}}\frac{\sin^2{x}\cdot\left(\sin^2{x}-3\sin{x}+2\right)}{\cos^2{x}\cdot\left(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\right)}=\\
=\lim_{x\to\frac{\pi}{2}}\frac{\sin^2{x}\cdot(\sin{x}-1)(\sin{x}-2)}{-(\sin{x}-1)(\sin{x}+1)\cdot\left(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\right)}=\\
=\lim_{x\to\frac{\pi}{2}}\frac{\sin^2{x}\cdot(\sin{x}-2)}{-(\sin{x}+1)\cdot\left(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\right)}
=\frac{1}{12}.
\]
Ответ:
\(\frac{1}{12}\)