Đáp án:
$L =\lim\limits_{x\to 0}\dfrac{(1-\cos2x)\sin2x}{x^2 + \tan x} =0$
Giải thích các bước giải:
$\begin{array}{l}\quad L =\lim\limits_{x\to 0}\dfrac{(1-\cos2x)\sin2x}{x^2 + \tan x}\\ \to L = \lim\limits_{x\to 0}\dfrac{2\sin x\cos^2x(1 - \cos2x)}{x^2\cos x + \sin x}\\ \to L = \lim\limits_{x\to 0}\dfrac{2\cos^2x(1 - \cos2x)}{x^2\dfrac{\cos x}{\sin x} + 1}\\ \to L = \dfrac{2\lim\limits_{x\to 0}\cos^2x.\lim\limits_{x\to 0}(1-\cos2x)}{\lim\limits_{x\to 0}\cos x\cdot\lim\limits_{x\to 0}\dfrac{x^2}{\sin x} +1}\\ \to L = \dfrac{2\lim\limits_{x\to 0}\cos^2x.\lim\limits_{x\to 0}(1-\cos2x)}{\lim\limits_{x\to 0}\cos x\cdot\lim\limits_{x\to 0}\dfrac{2x}{\cos x} +1}\\ \to L = \dfrac{2.\cos^20.[1 - \cos(2.0)]}{\cos0\cdot\dfrac{2.0}{\cos0} +1}\\ \to L = \dfrac{0}{1} = 0 \end{array}$
