Tính giới hạn $\Large A=\underset{x\rightarrow 0}{lim}\dfrac{\sqrt{4x+

Tính giới hạn $\Large A=\underset{x\rightarrow 0}{lim}\dfrac{\sqrt{4x+1}-\sqrt[3]{2x+1}}{x}$.

• A.

  $\Large +\infty$.

• B.

  $\Large -\infty$.

• C.

  $\Large \dfrac{4}{3}$.

• D.

  0.

Ta có $\Large A=\underset{x\rightarrow 0}{lim}\dfrac{\sqrt{4x+1}-1}{x}$ $\Large \underset{x\rightarrow 0}{lim}\dfrac{\sqrt[3]{2x+1}-1}{x}$

Mà $\Large \underset{x\rightarrow 0}{lim}\dfrac{\sqrt{4x+1}-1}{x}$ $\Large =\underset{x\rightarrow 0}{lim}\dfrac{4x}{x(\sqrt{4x+1}+1)}$ $\Large =\underset{x\rightarrow 0}{lim}\dfrac{4}{\sqrt{4x+1}+1}=2$

$\Large \underset{x\rightarrow 0}{lim}\dfrac{\sqrt[3]{2x+1}-1}{x}$ $\Large \underset{x\rightarrow 0}{lim}\dfrac{2x}{x\left[\sqrt[3]{(2x+1)^2}+\sqrt[3]{2x+1}+1\right]}=\dfrac{2}{3}$

Vậy $\Large A=2-\dfrac{2}{3}=\dfrac{4}{3}$. Vậy ta chọn đáp án C.