Question

1 пример = 25 баллов

Answer 1

Ответ:

$\lim\limits_{x \to 0}\dfrac{ln(cosx)}{ln(1+x^2)}=\lim\limits_{x \to 0}\dfrac{ln(1-1+cosx)}{ln(1+x^2)}=\lim\limits_{x \to 0}\dfrac{ln(1+(cosx-1))}{x^2}=\\\\\\=\lim\limits_{x \to 0}\dfrac{cosx-1}{x^2}=\lim\limits_{x \to 0}\dfrac{-2sin^2\frac{x}{2}}{x^2}=\lim\limits_{x \to 0}\dfrac{-2\cdot (\frac{x}{2})^2}{x^2}=\lim\limits_{x \to 0}\dfrac{-2x^2}{4\cdot x^2}=-\dfrac{1}{2}$

$\star \ \ ln(1+\alpha (x))\sim \alpha (x)\ \ ,\ \ esli\ \ \alpha (x)\to 0\\\\\star \ \ sin\alpha (x)\sim \alpha (x)\ \ ,\ \ esli\ \ \alpha (x)\to 0\\\\\star \ \ 1-cosx=2sin^2\frac{x}{2}$