$\lim\limits_{x\to 0} \dfrac{(x+2012).\sqrt[3]{1-2x}-2012\sqrt{4x+1}}{x}$

Đáp án:

$-\dfrac{16093}{3}$

Giải thích các bước giải:

$L = \lim\limits_{x\to 0} \dfrac{(x+2012).\sqrt[3]{1-2x}-2012\sqrt{4x+1}}{x}$

$= 2012\lim\limits_{x\to 0}\dfrac{\sqrt[3]{1-2x} - \sqrt{4x+1}}{x} + \lim\limits_{x\to 0}\sqrt[3]{1-2x}$

Đặt $\begin{cases}L_1 = \lim\limits_{x\to 0}\dfrac{\sqrt[3]{1-2x} - \sqrt{4x+1}}{x}\\L_2 = \lim\limits_{x\to 0}\sqrt[3]{1-2x}\end{cases}$

Ta được:

$L = 2012L_1 + L_2$

$\bullet\ \ L_2 = \lim\limits_{x\to 0}\sqrt[3]{1-2x}$

$\Leftrightarrow L_2 = \sqrt[3]{1-2.0}$

$\Leftrightarrow L_2 = 1$

$\bullet\ \ L_1 = \lim\limits_{x\to 0}\dfrac{\sqrt[3]{1-2x} - \sqrt{4x+1}}{x}$

$\Leftrightarrow L_1 = \lim\limits_{x\to 0}\dfrac{\left(\sqrt[3]{1-2x} - 1\right) - \left(\sqrt{4x+1} - 1\right)}{x}$

$\Leftrightarrow L_1 = \lim\limits_{x\to 0}\dfrac{\dfrac{-2x}{\sqrt[3]{(1-2x)^2} + \sqrt[3]{1-2x} + 1} - \dfrac{4x}{\sqrt{4x+1} + 1}}{x}$

$\Leftrightarrow L_1 = \lim\limits_{x\to 0}\left(\dfrac{-2}{\sqrt[3]{(1-2x)^2} + \sqrt[3]{1-2x} + 1} - \dfrac{4}{\sqrt{4x+1} + 1}\right)$

$\Leftrightarrow L_1 =\dfrac{-2}{\sqrt[3]{(1-2.0)^2} + \sqrt[3]{1-2.0} + 1} - \dfrac{4}{\sqrt{4.0+1} + 1}$

$\Leftrightarrow L_1 = - \dfrac23 - 2$

$\Leftrightarrow L_1 = -\dfrac83$

Khi đó:

$L = 2012L_1 + L_2 = 2012\cdot \left(-\dfrac83\right) + 1 =-\dfrac{16093}{3}$