Найти предел:
\(
\lim_{x\to\frac{\pi}{6}} \large \frac{1-2\sin{x}}{\cos{3x}}.
\)

Решение

\( \displaylines{
\lim_{x\to\frac{\pi}{6}} \frac{1-2\sin{x}}{\cos{3x}}
= \lim_{x\to\frac{\pi}{6}} \frac{-2\cos{x}}{-3\sin{3x}}
=\\
= \lim_{x\to\frac{\pi}{6}} \frac{2\cos{\frac{\pi}{6}}}{3\sin{\frac{\pi}{6}}}
= \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}
}\)

Ответ: \( \large \frac{1}{\sqrt{3}} \).