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[Mmm4]

lim [tex]\frac{ \sqrt[3]{tgx}-1}{2 sin^{ 2 }x - 1 }[/tex]
x->[tex]\frac{ \pi }{4}[/tex]

[KOPMOPAH]

$$\lim_{x \to \frac{ \pi }{4}} \frac{ \sqrt[3]{\tg x}-1}{2 \sin^{ 2 }x - 1 }=\lim_{x \to \frac{ \pi }{4}} \frac{ \sqrt[3]{\tg x}-1}{ \sin^ 2 x - \cos^2 x }=\lim_{x \to \frac{ \pi }{4}} \frac{( \sqrt[3]{\tg x}-1)\left(\sqrt[3]{\tg ^2x}+\sqrt[3]{\tg x}+1\right)}{ (\sin x - \cos x)(\sin x + \cos x) (\sqrt[3]{\tg ^2x}+\sqrt[3]{\tg x}+1)}=$$
$$=\lim_{x \to \frac{ \pi }{4}}\frac{\tg x-1}{ (\sin x - \cos x)(\sin x + \cos x) (\sqrt[3]{\tg ^2x}+\sqrt[3]{\tg x}+1)}=\lim_{x \to \frac{ \pi }{4}}\frac{\frac {\cancel{\sin x-\cos x}}{\cos x}}{ \cancel{(\sin x - \cos x)}(\sin x + \cos x) (\sqrt[3]{\tg ^2x}+\sqrt[3]{\tg x}+1)}=$$
Отървахме се от неопределеността $ \frac{0}{0}$ и действаме $$=\lim_{x \to \frac{ \pi }{4}}\frac{1}{\cos x(\sin x + \cos x) (\sqrt[3]{\tg ^2x}+\sqrt[3]{\tg x}+1)}=\cdots$$

[Mmm4]

Благодаря!