D

\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits_{x\rightarrow-1}\left[\dfrac{x+1}{\left(\sqrt[3]{x}\right)^2-\sqrt[3]{x}+1}:\dfrac{x^2-1}{\sqrt{x^2+3}+2}\right]\\ =\lim\limits_{x\rightarrow-1}\left[\dfrac{x+1}{\left(\sqrt[3]{x}\right)^2-\sqrt[3]{x}+1}:\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+3}+2}\right]\\ =\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x^2+3}+2}{\left[\left(\sqrt[3]{x}\right)^2-\sqrt[3]{x}+1\right]\left(x-1\right)}=-\dfrac{2}{3}\)

\(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{3x-2}}{x^4-4}=0\)

\(\lim\limits_{x\rightarrow1}\dfrac{x+2-\sqrt{x+8}}{\sqrt{x+3}+x-3}=\lim\limits_{x\rightarrow1}\left(\dfrac{x^2+3x-4}{x+2+\sqrt{x+8}}:\dfrac{-x^2+7x-6}{\sqrt{x+3}+3-x}\right)\\ =\lim\limits_{x\rightarrow1}\left[\dfrac{\left(x-1\right)\left(x+4\right)}{x+2+\sqrt{x+8}}:\dfrac{\left(x-1\right)\left(6-x\right)}{\sqrt{x+3}+3-x}\right]\\ =\lim\limits_{x\rightarrow1}\dfrac{\left(x+4\right)\left(\sqrt{x+3}+3-x\right)}{\left(6-x\right)\left(x+2+\sqrt{x+8}\right)}=\dfrac{2}{3}\)

Sai.

\(\lim\limits_{x\rightarrow1^-}\dfrac{x\sqrt{1-x}}{2\sqrt{1-x}+1-x}=\lim\limits_{x\rightarrow1^-}\dfrac{x\sqrt{1-x}}{\sqrt{1-x}\left(2+\sqrt{1-x}\right)}\\ =\lim\limits_{x\rightarrow1^-}\dfrac{x}{2+\sqrt{1-x}}=\dfrac{1}{2}\)

Không tồn tại.

\(\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{\left(1-x\right)\left(2-x\right)}}{\left(1-x\right)\left(4-x\right)}\\ =\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{2-x}}{\left(4-x\right)\sqrt{1-x}}\)

\(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1^-}\sqrt{2-x}=1>0\\\lim\limits_{x\rightarrow1^-}\left(4-x\right)\sqrt{1-x}=0\\x< 1\rightarrow\left(4-x\right)\sqrt{1-x}>0\end{matrix}\right.\\ \rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{2-x}}{\left(4-x\right)\sqrt{1-x}}=+\infty\) 

Không tồn tại.

Sai.