Tìm giới hạn A= lim (x->0)( căn (4x+1)- căn bậc 3 (2x+1))/x

* Hướng dẫn giải

Chọn C

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

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

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

Vậy $A=2-\frac{2}{3}=\frac{4}{3}$