Question

Помогите решить (найти предел функции)

Answer 1

$\lim_{x \to 0} ( \frac{ \cos(x) }{ \sin {}^{2} (x) } - \cot {}^{2} (x)) = \lim_{x \to0}( \frac{ \cos(x) }{ \sin {}^{2} (x) } - \frac{ \cos {}^{2} (x) }{ \sin {}^{2} (x) } ) = \lim_{x \to 0} \frac{ \cos(x) - \cos {}^{2} (x) }{ \sin {}^{2} (x) } = \lim_{x \to 0} \frac{( \cos(x))' - ( \cos {}^{2} (x)) ' }{( \sin {}^{2} (x))'} = \lim_{x \to 0} \frac{ - \sin(x) + \sin(2x) }{ \sin(2x) } = \lim_{x \to 0} \frac{ - \sin(x) + 2 \sin(x) \cos(x) }{2 \sin(x) \cos(x) } = \lim_{x \to 0} \frac{ - \sin(x)(1 - 2 \cos(x) ) }{2 \sin(x) \cos(x) } = \lim_{x \to 0} \frac{2 \cos(x) - 1}{2 \cos(x) } = \frac{2 \cos(0) - 1 }{2 \cos(0) } = \frac{1}{2}$

Ответ: 0.5