Tìm giới hạn $\Large E=\underset{x\rightarrow 7}{lim}\dfrac{\sqrt[3]{4

Tìm giới hạn $\Large E=\underset{x\rightarrow 7}{lim}\dfrac{\sqrt[3]{4x-1}-\sqrt{x+2}}{\sqrt[4]{2x+2}-2}$:

• A.

  $\Large +\infty$.

• B.

  $\Large -\infty$.

• C.

  $\Large \dfrac{-8}{27}$.

• D.

  1.

Chọn C.

Ta có: $\Large E=\underset{x\rightarrow 7}{lim}\dfrac{\sqrt[3]{4x-1}-\sqrt{x+2}}{\sqrt[4]{2x+2}-2}$ $\Large =\underset{x\rightarrow 7}{lim}\dfrac{\sqrt[3]{4x-1}-3}{\sqrt[4]{2x+2}-2}-\underset{x\rightarrow 7}{lim}\dfrac{\sqrt{x+2}-3}{\sqrt[4]{2x+2}-2}$ $\Large =A-B$

$\Large A=\underset{x\rightarrow 7}{lim}\dfrac{\sqrt[3]{4x-1}-3}{\sqrt[4]{2x+2}-2}$ $\Large =\underset{x\rightarrow 7}{lim}\dfrac{\left(\sqrt[4]{2x+2}+2\right)\left(\sqrt[4]{(2x+2)^2}+4\right)}{\left(\sqrt[3]{(4x-1)^2}+3\sqrt[3]{4x-1}+9\right)}=\dfrac{64}{27}$

$\Large B==\underset{x\rightarrow 7}{lim}\dfrac{\sqrt{x+2}-3}{\sqrt[4]{2x+2}-2}$ $\Large =\underset{x\rightarrow 7}{lim}\dfrac{\left(\sqrt[4]{2x+2}+2\right)\left(\sqrt[4]{(2x+2)^2}+4\right)}{2\left(\sqrt{x+2}+2\right)}=\dfrac{8}{3}$

$\Large E=A-B=\dfrac{64}{27}-\dfrac{8}{3}=\dfrac{-8}{27}$.