Đáp án:
$\dfrac52$
Giải thích các bước giải:
$\lim\limits_{x\to 0}\dfrac{\sqrt{\tan6x +1} -\sqrt{\sin x +1}}{x}$
$= \lim\limits_{x\to 0}\dfrac{(\sqrt{\tan6x +1} -\sqrt{\sin x +1})(\sqrt{\tan6x +1} +\sqrt{\sin x+1})}{x(\sqrt{\tan6x +1} +\sqrt{\sin x +1})}$
$=\lim\limits_{x\to 0}\dfrac{\tan6x -\sin x}{x(\sqrt{\tan6x +1} +\sqrt{\sin x +1})}$
$=\lim\limits_{x\to 0}\dfrac{\sin6x -\cos6x\sin x}{x\cos6x(\sqrt{\tan6x +1} +\sqrt{\sin x +1})}$
$=\lim\limits_{x\to 0}\dfrac{\sin6x -\cos6x\sin x}{x}\cdot\lim\limits_{x\to 0}\dfrac{1}{\cos6x(\sqrt{\tan6x +1} +\sqrt{\sin x +1})}$
$=\left(\lim\limits_{x\to 0}\dfrac{\sin6x}{x} - \lim\limits_{x\to 0}\dfrac{\sin x}{x}\cdot\lim\limits_{x\to 0}\cos6x\right)\cdot\dfrac{1}{\cos0(\sqrt{\tan0 +1} +\sqrt{\sin0 +1})}$
$=\dfrac12\cdot\left(\lim\limits_{x\to 0}\dfrac{\sin6x}{x} - \lim\limits_{x\to 0}\dfrac{\sin x}{x}\cdot\lim\limits_{x\to 0}\cos6x\right)$
$=\dfrac12\cdot\left(\lim\limits_{x\to 0}\dfrac{6\cos6x}{1} - \lim\limits_{x\to 0}\dfrac{\cos x}{1}\cdot\cos0\right)$
$=\dfrac12\cdot(6\cos0 - \cos0)\cdot 1$
$= \dfrac52$
