Question

302) по методу первого замечательного предела

Answer 1

$\lim\limits _{x \to \pi /6}\, \frac{1-2sinx}{\pi /6-x}=\lim\limits _{x \to \pi /6}\frac{2\cdot (\frac{1}{2}-sinx)}{\pi /6-x}=\lim\limits _{x \to \pi /6}\frac{2\cdot (sin\frac{\pi}{6}-sinx)}{\pi /6-x}=\\\\=\lim\limits _{x \to \pi /6}\frac{2\cdot 2\, sin(\frac{\pi}{12}-\frac{x}{2})\cdot cos(\frac{\pi}{12}+\frac{x}{2})}{\pi /6-x}=\\\\=\Big[\; sin(\frac{\pi}{12}-\frac{x}{2})\sim (\frac{\pi}{12}-\frac{x}{2})\; ,\; tak\; kak\; \; (\frac{\pi}{12}-\frac{x}{2})\to 0\; ,\; esli\; \; x\to \frac{\pi}{6}\; \Big]=$

$=\lim\limits _{x \to \pi /6}\frac{2\cdot 2\, (\frac{\pi}{12}-\frac{x}{2})\cdot cos(\frac{\pi}{12}+\frac{x}{2})}{\frac{\pi}{6}-x}=\lim\limits _{x \to \pi /6}\frac{2\cdot (\frac{\pi}{6}-x)\cdot cos(\frac{\pi}{12}+\frac{x}{2})}{\frac{\pi}{6}-x}=\\\\=2\cdot \lim\limits _{x \to \pi /6}\; cos(\frac{\pi}{12}+\frac{x}{2})=2\cdot cos(\frac{\pi}{12}+\frac{\pi}{12})=2\cdot cos\frac{\pi}{6}=2\cdot \frac{\sqrt3}{2}=\sqrt3$

$P.S.\; \; \; \lim\limits _{\alpha (x) \to 0}\frac{sin\,\alpha (x)}{\alpha (x)}=1\; \; \Rightarrow \; \; \; sin\,\alpha (x)\sim \alpha (x)$

Answer 2

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