Tính giới hạn $L=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{{{x}^{2}}}$

Đề bài: (Câu hỏi của bạn Suka Mori trên facebook)
Tính giới hạn $L=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{{{x}^{2}}}$
Giải:
Ta có: $$L=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \sqrt{1+2x}-\left( x+1 \right) \right)-\left( \sqrt[3]{1+3x}-\left( x+1 \right) \right)}{{{x}^{2}}}$$

$$\begin{align}   & =\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+2x}-\left( x+1 \right)}{{{x}^{2}}}-\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt[3]{1+3x}-\left( x+1 \right)}{{{x}^{2}}} \\  & =\underset{x\to 0}{\mathop{\lim }}\,\frac{-{{x}^{2}}}{{{x}^{2}}\left( \sqrt{1+2x}+x+1 \right)}-\underset{x\to 0}{\mathop{\lim }}\,\frac{-{{x}^{3}}-3{{x}^{2}}}{{{x}^{2}}\left[ \sqrt[3]{{{\left( 1+3x \right)}^{2}}}+\sqrt[3]{1+3x}\left( x+1 \right)+{{\left( x+1 \right)}^{2}} \right]} \\  & =\underset{x\to 0}{\mathop{\lim }}\,\frac{-1}{\sqrt{1+2x}+x+1}+\underset{x\to 0}{\mathop{\lim }}\,\frac{x+3}{\sqrt[3]{{{\left( 1+3x \right)}^{2}}}+\sqrt[3]{1+3x}\left( x+1 \right)+{{\left( x+1 \right)}^{2}}}=\frac{3}{1+1+1}-\frac{1}{1+1}=\frac{1}{2} \\ \end{align}$$    

Vậy $L=\frac{1}{2}$