Giải thích các bước giải:
$\lim_{x\to 0}\dfrac{\sqrt{1-2x-5x^2}-\sqrt[3]{1+3x-x^2}}{x^2}$
$=\lim_{x\to 0}\dfrac{(\sqrt{1-2x-5x^2}-1)+(1-\sqrt[3]{1+3x-x^2})}{x^2}$
$=\lim_{x\to 0}\dfrac{\dfrac{1-2x-5x^2-1}{\sqrt{1-2x-5x^2}+1}+\dfrac{1-(1+3x-x^2)}{1+\sqrt[3]{1+3x-x^2}+\sqrt[3]{1+3x-x^2}^2}}{x^2}$
$=\lim_{x\to 0}\dfrac{\dfrac{-2x-5x^2}{\sqrt{1-2x-5x^2}+1}+\dfrac{x^2-3x}{1+\sqrt[3]{1+3x-x^2}+\sqrt[3]{1+3x-x^2}^2}}{x^2}$
$=\lim_{x\to 0}\dfrac{\dfrac{-2-5x}{\sqrt{1-2x-5x^2}+1}+\dfrac{x-3}{1+\sqrt[3]{1+3x-x^2}+\sqrt[3]{1+3x-x^2}^2}}{x}$
$=\dfrac{\dfrac{-2-5.0}{\sqrt{1-2.0-5.0^2}+1}+\dfrac{0-3}{1+\sqrt[3]{1+3.0-0^2}+\sqrt[3]{1+3.0-0^2}^2}}{0}$
$=\dfrac{-1-1}{0}$
$=-\infty$
