Giải thích các bước giải:
Ta có:
$\lim_{x\to0}\dfrac{\sqrt[3]{6x+1}-\sqrt{4x+1}}{x^2}$
$=\lim_{x\to0}\dfrac{(\sqrt[3]{6x+1}-(2x+1))-(\sqrt{4x+1}-(2x+1))}{x^2}$
$=\lim_{x\to0}\dfrac{\dfrac{6x+1-(2x+1)^3}{(\sqrt[3]{6x+1})^2+\sqrt[3]{6x+1}(2x+1)+(2x+1)^2}-\dfrac{4x+1-(2x+1)^2}{\sqrt{4x+1}+(2x+1)}}{x^2}$
$=\lim_{x\to0}\dfrac{\dfrac{-8x^3-12x^2}{(\sqrt[3]{6x+1})^2+\sqrt[3]{6x+1}(2x+1)+(2x+1)^2}-\dfrac{-4x^2}{\sqrt{4x+1}+(2x+1)}}{x^2}$
$=\lim_{x\to0}\dfrac{-8x-12}{(\sqrt[3]{6x+1})^2+\sqrt[3]{6x+1}(2x+1)+(2x+1)^2}-\dfrac{-4}{\sqrt{4x+1}+(2x+1)}$
$=\dfrac{-8\cdot 0-12}{(\sqrt[3]{6\cdot 0+1})^2+\sqrt[3]{6\cdot 0+1}(2\cdot 0+1)+(2\cdot 0+1)^2}-\dfrac{-4}{\sqrt{4\cdot 0+1}+(2\cdot 0+1)}$
$=-2$
