Giải thích các bước giải:
$L=\lim_{x\to 0}\dfrac{(x^2+2012)\sqrt[3]{1-2x}-2012\sqrt{4x+1}}{x}$
$\to L=\lim_{x\to 0}x\sqrt[3]{1-2x}+\dfrac{2012(\sqrt[3]{1-2x}-\sqrt{4x+1})}{x}$
$\to L=\lim_{x\to 0}x\sqrt[3]{1-2x}+\dfrac{2012((\sqrt[3]{1-2x}-1)-(\sqrt{4x+1}-1))}{x}$
$\to L=\lim_{x\to 0}x\sqrt[3]{1-2x}+\dfrac{2012(\dfrac{1-2x-1}{\sqrt[3]{1-2x}^2+\sqrt[3]{1-2x}+1}-\dfrac{4x+1-1}{\sqrt{4x+1}+1})}{x}$
$\to L=\lim_{x\to 0}x\sqrt[3]{1-2x}+\dfrac{2012(\dfrac{-2x}{\sqrt[3]{1-2x}^2+\sqrt[3]{1-2x}+1}-\dfrac{4x}{\sqrt{4x+1}+1})}{x}$
$\to L=\lim_{x\to 0}x\sqrt[3]{1-2x}+2012(\dfrac{-2}{\sqrt[3]{1-2x}^2+\sqrt[3]{1-2x}+1}-\dfrac{4}{\sqrt{4x+1}+1})$
$\to L=0.\sqrt[3]{1-2.0}+2012(\dfrac{-2}{\sqrt[3]{1-2.0}^2+\sqrt[3]{1-2.0}+1}-\dfrac{4}{\sqrt{4.0+1}+1})$
$\to L=-\dfrac{16096}{3}$
