sin3x...

August 2021 1 45 Report

Помогите, пожалуйста, вычислить предел,
используя первый замечательный предел:
limx→0 (1 - cosx)/sin3x

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NNNLLL54

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$\lim\limits _{x\to 0} \frac{1-cosx}{sin3x} = \lim\limits _{x\to 0}\frac{2sin^2\frac{x}{2}}{sin3x} = 2\cdot \lim\limits _{x \to 0} \Big (\underbrace {\frac{sin\frac{x}{2}}{\frac{x}{2}}}_{1}\cdot \underbrace {\frac{sin\frac{x}{2}}{\frac{x}{2}}}_{1}\cdot \underbrace {\frac{3x}{sin3x}}_{1}\cdot \frac{(\frac{x}{2})^2}{3x}\Big )=\\\\=2\cdot \lim\limits _{n \to \infty} \frac{(\frac{x}{2})^2}{3}=2\cdot \lim\limits _{x \to 0} \frac{x^2}{12x} =2\cdot \lim\limits _{x \to 0} \frac{x}{12} =2\cdot 0=0$

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