Đáp án:
$\lim\limits_{x\to 0}\dfrac{\sqrt{1 - x^2} -1}{2 - \sqrt{4 - \tan^2x}} = -2$
Giải thích các bước giải:
$\quad \lim\limits_{x\to 0}\dfrac{\sqrt{1 - x^2} -1}{2 - \sqrt{4 - \tan^2x}}$
$= \lim\limits_{x\to 0}\dfrac{\left(\sqrt{1 - x^2} -1\right)\left(\sqrt{1 - x^2} +1\right)\left(2 + \sqrt{4 -\tan^2x}\right)}{\left(2 - \sqrt{4 - \tan^2x}\right)\left(2 + \sqrt{4 -\tan^2x}\right)\left(\sqrt{1 - x^2} +1\right)}$
$= \lim\limits_{x\to 0}\dfrac{-x^2\left(2 + \sqrt{4 - \tan^2x}\right)}{\tan^2x\left(\sqrt{1 - x^2 } + 1\right)}$
$= \lim\limits_{x\to 0}\dfrac{-x^2}{\tan^2x}\cdot \lim\limits_{x\to 0}\dfrac{2 + \sqrt{4 - \tan^2x}}{\sqrt{1 - x^2 } + 1}$
$= \lim\limits_{x\to 0}\dfrac{-x^2}{x^2}\cdot \dfrac{2 + \sqrt{4 - \tan^20}}{\sqrt{1 - 0^2 } + 1}\qquad (\tan x\sim x)$
$= - 1 \cdot 2$
$= -2$
