Найти предел lim x- больше 0 √(1+xsin8x) - 1 / tg^2(2x)

[latex] \lim_{x \to 0} \frac{ \sqrt{1+xsin8x}-1 }{tg^22x} = \lim_{x \to 0} \frac{ \frac{sin8x+8xcos8x}{2\sqrt{1+xsin8x}} }{ \frac{4tg2x}{cos^22x} } = \lim_{x \to 0} \frac{ \frac{sin8x+8xcos8x}{2} }{ \frac{4tg2x}{1} } [/latex][latex]= \lim_{x \to 0} \frac{sin8x+8xcos8x}{8tg2x} = \lim_{x \to 0} \frac{16cos8x - 64xsin8x}{ \frac{16}{cos^22x} } = \frac{16}{16} = 1[/latex]