Tính lim x tiến tới 3 của căn bậc hai của x+1 - căn bậc ba của 2x+2

* Hướng dẫn giải

$\begin{array}{l}\underset{x\to 3}{\lim }\frac{\sqrt{x+1}-\sqrt[3]{2x+\text{}2}}{x-3}=\underset{x\to 3}{\lim }\frac{\sqrt{x+1}-2+\text{\hspace{0.17em}\hspace{0.17em}}2-\sqrt[3]{2x+\text{}2}}{x-3}\\ =\underset{x\to 3}{\lim }\frac{(\sqrt{x+1}-2).(\sqrt{x+1}+2)}{(x-3).(\sqrt{x+1}+2)}+\text{\hspace{0.17em}\hspace{0.17em}}\underset{x\to 3}{\lim }\frac{\text{\hspace{0.17em}\hspace{0.17em}}(2-\sqrt[3]{2x+\text{}2}).\left(4+2\sqrt[3]{2x+\text{}2}+\text{}{\sqrt[3]{2x+\text{}2}}^2\right)}{(x-3).\left(4+2\sqrt[3]{2x+\text{}2}+\text{}{\sqrt[3]{2x+\text{}2}}^2\right)}\\ =\underset{x\to 3}{\lim }\frac{x+\text{}1-4}{(x-3).(\sqrt{x+1}+2)}+\text{\hspace{0.17em}}\underset{x\to 3}{\lim }\frac{\text{\hspace{0.17em}\hspace{0.17em}}8-(2x+\text{}2)}{(x-3).\left(4+2\sqrt[3]{2x+\text{}2}+\text{}{\sqrt[3]{2x+\text{}2}}^2\right)}\\ =\underset{x\to 3}{\lim }\frac{x-3}{(x-3).(\sqrt{x+1}+2)}+\text{\hspace{0.17em}}\underset{x\to 3}{\lim }\frac{\text{\hspace{0.17em}\hspace{0.17em}}-2(x-3)}{(x-3).\left(4+2\sqrt[3]{2x+\text{}2}+\text{}{\sqrt[3]{2x+\text{}2}}^2\right)}\\ =\underset{x\to 3}{\lim }\frac{1}{\sqrt{x+1}+2}+\text{\hspace{0.17em}}\underset{x\to 3}{\lim }\frac{\text{\hspace{0.17em}\hspace{0.17em}}-2}{\left(4+2\sqrt[3]{2x+\text{}2}+\text{}{\sqrt[3]{2x+\text{}2}}^2\right)}\\ =\frac{1}{\sqrt{3+1}+2}+\text{\hspace{0.17em}}\frac{-2}{4+\text{}4\text{}+4}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\end{array}$

Chọn B