Question

[Quỳnh Anh]

Tìm giới hạn: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+2x}\sqrt[3]{1+3x}-1}{x}\)

Answer 1

[nguyễn thị hương giang]

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:

\(^{lim}_{x\rightarrow0}\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}=1+\dfrac{2}{3}=\dfrac{5}{3}\)