Lim стремится к 0 sin5x/tg2x

Решите задачу:

$\lim_{n \to \inft0} \frac{sin5x}{tg2x} = \lim_{n \to \inft0} \frac{sin5x}{ \frac{sin2x}{cos2x} } = \lim_{n \to \inft0} cos2x\frac{sin5x}{sin2x } = \\ \\ = \lim_{n \to \inft0} cos2x * \lim_{n \to \inft0} \frac{sin5x}{sin2x } = 1*\lim_{n \to \inft0} \frac{sin5x}{sin2x } = \\ \\ = \lim_{n \to \inft0} \frac{ \frac{sin5x}{x} }{ \frac{sin2x}{x} } =\lim_{n \to \inft0} \frac{ \frac{5 sin5x}{5x} }{ \frac{2 sin2x}{2x} } = \frac{5}{2} \lim_{n \to \inft0} \frac{ \frac{sin5x}{5x} }{ \frac{sin2x}{2x} } =$

$= \frac{5}{2} \frac{\lim_{n \to \inft0} \frac{sin5x}{5x} }{\lim_{n \to \inft0} \frac{sin2x}{2x} } = \frac{5}{2} \frac{1}{1} = \frac{5}{2}$