$\underset{x \to 0}{lim} \dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}$

Tính giới hạn    $\underset{x \to 0}{lim} \dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}$

$\underset{x \to 0}{lim} \dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}$

$= \underset{x \to 0}{lim} \dfrac{\sqrt{1+4x} - (2x+1) +(2x+1)-\sqrt[3]{1+6x}}{x^2}$

$= \underset{x \to 0}{lim} \dfrac{ \dfrac{-4x^2}{\sqrt{1+4x} + (2x+1)} + \dfrac{8x^3+12x^2}{(2x+1)^2+(2x+1)\sqrt[3]{1+6x}+(\sqrt[3]{1+6x})^2}}      {x^2}$

$= \underset{x \to 0}{lim} \left ( \dfrac{-4}{\sqrt{1+4x} + (2x+1)} + \dfrac{8x+12}{(2x+1)^2+(2x+1)\sqrt[3]{1+6x}+(\sqrt[3]{1+6x})^2} \right )$

$=2$