Tính lim từ x đến 0 căn bậc hai (1+2x). căn bậc ba (1+3x). căn bậc bốn (1+4x)-1/x

* Hướng dẫn giải

Ta có:

$\begin{array}{l}\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1\\ =\sqrt{1+2x}-\sqrt{1+2x}+\sqrt{1+2x}.\sqrt[3]{1+3x}-\sqrt{1+2x}.\sqrt[3]{1+3x}+\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1\\ =\left(\sqrt{1+2x}-1\right)+\sqrt{1+2x}\left(\sqrt[3]{1+3x}-1\right)+\sqrt{1+2x}.\sqrt[3]{1+3x}.\left(\sqrt[4]{1+4x}-1\right)\end{array}$

$\begin{array}{l}\Rightarrow \underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}\\ =\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+2x}-1}{x}\right)+\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\frac{\sqrt[3]{1+3x}-1}{x}\right)+\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\sqrt[4]{1+4x}-1}{x}\right)\end{array}$

Tính:

$\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+2x}-1}{x}\right)=\underset{x\to 0}{\lim }\frac{\left(\sqrt{1+2x}-1\right)\left(\sqrt{1+2x}+1\right)}{x\left(\sqrt{1+2x}+1\right)}=\underset{x\to 0}{\lim }\frac{2x}{x\left(\sqrt{1+2x}+1\right)}=\underset{x\to 0}{\lim }\frac{2}{\sqrt{1+2x}+1}=\frac{2}{1+1}=1$

$\begin{array}{l}\underset{x\to 0}{lim}\left(\sqrt{1+2x}.\frac{\sqrt[3]{1+3x}-1}{x}\right)\\ =\underset{x\to 0}{lim}\left(\sqrt{1+2x}.\frac{\left(\sqrt[3]{1+3x}-1\right)\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}{x.\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)\\ =\underset{x\to 0}{lim}\left(\sqrt{1+2x}.\frac{3x}{x.\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)\\ =\underset{x\to 0}{lim}\left(\frac{3\sqrt{1+2x}}{\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)=\frac{3.1}{1+1+1}=3\end{array}$

$\underset{x\to 0}{lim}\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\sqrt[4]{1+4x}-1}{x}\right)$

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

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

$=\underset{x\to 0}{lim}\frac{4\sqrt{1+2x}.\sqrt[3]{1+3x}}{\left[{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1\right]}=\frac{\mathrm{4.1.1}}{1+1+1+1}=1$

Vậy $\underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}=1+1+1=3$