Задача №2030
Условие
Найти предел \(\lim_{x\to{0}}\frac{x\ctg{x}-1}{x^2}\).
Решение
\[
\lim_{x\to{0}}\frac{x\ctg{x}-1}{x^2}
=\left[\frac{0}{0}\right]
=\lim_{x\to{0}}\frac{\left(x\ctg{x}-1\right)'}{\left(x^2\right)'}
=\lim_{x\to{0}}\frac{\ctg{x}-\frac{x}{\sin^2{x}}}{2x}
=\lim_{x\to{0}}\frac{\cos{x}\sin{x}-x}{2x\sin^2{x}}=\\
=\lim_{x\to{0}}\frac{\frac{1}{2}\sin{2x}-x}{2x\sin^2{x}}
=\left[\frac{0}{0}\right]
=\lim_{x\to{0}}\frac{\left(\frac{1}{2}\sin{2x}-x\right)'}{\left(2x\sin^2{x}\right)'}
=\lim_{x\to{0}}\frac{\cos{2x}-1}{\sin^2{x}+2x\sin{x}\cos{x}}=\\
=\lim_{x\to{0}}\frac{-\sin^2{x}}{\sin^2{x}+2x\sin{x}\cos{x}}
=\lim_{x\to{0}}\frac{-\frac{\sin{x}}{x}}{\frac{\sin{x}}{x}+2\cos{x}}
=-\frac{1}{3}.
\]
Ответ:
\(-\frac{1}{3}\)