Ta xét:
\(\sqrt{1+2x}\cdot\sqrt[3]{1+3x}-1\)
\(=\sqrt{1+2x}-\sqrt{1+2x}+\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1\)
\(=\left(\sqrt{1+2x}-1\right)+\sqrt{1+2x}\cdot\left(\sqrt[3]{1+2x}-1\right)\)
Xét giới hạn trên:
\(\Rightarrow^{lim}_{x\rightarrow0}\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}\)
\(=^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}-1}{x}\right)+^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}\cdot\left(\sqrt[3]{1+2x}-1\right)}{3}\right)\)
Tính giới hạn từng thành phần:
* \(^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}-1}{x}\right)=^{lim}_{x\rightarrow0}\left(\dfrac{1+2x-1}{x\left(\sqrt{1+2x}+1\right)}\right)\)
\(=^{lim}_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+2x}+1}\right)=\dfrac{2}{\sqrt{1+2\cdot0}+1}=1\left(1\right)\)
* \(^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}\right)\)
\(=^{lim}_{x\rightarrow0}\left(\sqrt{1+2x}\cdot\dfrac{1+2x-1}{x\left(\left(\sqrt[3]{1+2x}\right)^2+\sqrt[3]{1+2x}+1\right)}\right)\)
\(=^{lim}_{x\rightarrow0}\left(\sqrt{1+2x}\cdot\dfrac{2}{\left(\sqrt[3]{1+2x}\right)^2+\sqrt[3]{1+2x}+1}\right)\)
\(=\sqrt{1+2\cdot0}\cdot\dfrac{2}{(\sqrt[3]{1+2\cdot0})^2+\sqrt[3]{1+2\cdot0}+1}\)
\(=\dfrac{2}{3}\left(2\right)\)
Lấy \(\left(1\right)+\left(2\right)\) ta được:
\(^{lim}_{x\rightarrow0}\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}=1+\dfrac{2}{3}=\dfrac{5}{3}\)
