Phương pháp giải:

Tính giới hạn của các hàm số ở từng đáp án.

Lời giải chi tiết:

\(\eqalign{  & \mathop {\lim }\limits_{x \to 0} {{\sqrt {x + 1}  - \root 3 \of {x + 1} } \over x} = \mathop {\lim }\limits_{x \to 0} \left( {{{\sqrt {x + 1}  - 1} \over x} - {{\root 3 \of {x + 1}  - 1} \over x}} \right)  \cr   &  = \mathop {\lim }\limits_{x \to 0} {{\sqrt {x + 1}  - 1} \over x} - \mathop {\lim }\limits_{x \to 0} {{\root 3 \of {x + 1}  - 1} \over x}  \cr   &  = \mathop {\lim }\limits_{x \to 0} {{x + 1 - 1} \over {x\left( {\sqrt {x + 1}  + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} {{x + 1 - 1} \over {x\left( {{{\root 3 \of {x + 1} }^2} + \root 3 \of {x + 1}  + 1} \right)}}  \cr   &  = \mathop {\lim }\limits_{x \to 0} {1 \over {\sqrt {x + 1}  + 1}} - \mathop {\lim }\limits_{x \to 0} {1 \over {{{\root 3 \of {x + 1} }^2} + \root 3 \of {x + 1}  + 1}}  \cr   &  = {1 \over {1 + 1}} - {1 \over {1 + 1 + 1}} = {1 \over 6} \cr} \)

\( \Rightarrow \) Đáp án A sai.

\(\mathop {\lim }\limits_{x \to 1} {{\sqrt {5 - x}  - 2} \over {\sqrt {2 - x}  - 1}} = \mathop {\lim }\limits_{x \to 1} {{\left( {5 - x - 4} \right)\left( {\sqrt {2 - x}  + 1} \right)} \over {\left( {2 - x - 1} \right)\left( {\sqrt {5 - x}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} {{\left( {1 - x} \right)\left( {\sqrt {2 - x}  + 1} \right)} \over {\left( {1 - x} \right)\left( {\sqrt {5 - x}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} {{\sqrt {2 - x}  + 1} \over {\sqrt {5 - x}  + 2}} = {{1 + 1} \over {2 + 2}} = {1 \over 2}\)

\( \Rightarrow \) Đáp án B sai.

\(\eqalign{  & \mathop {\lim }\limits_{x \to 1} {{\root 3 \of x  - \sqrt x } \over {{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {{{\root 3 \of x  - 1} \over {{x^2} - 1}} - {{\sqrt x  - 1} \over {{x^2} - 1}}} \right) = \mathop {\lim }\limits_{x \to 1} {{\root 3 \of x  - 1} \over {{x^2} - 1}} - \mathop {\lim }\limits_{x \to 1} {{\sqrt x  - 1} \over {{x^2} - 1}}  \cr   &  = \mathop {\lim }\limits_{x \to 1} {{x - 1} \over {\left( {x - 1} \right)\left( {x + 1} \right)\left( {{{\root 3 \of x }^2} + \root 3 \of x  + 1} \right)}} - \mathop {\lim }\limits_{x \to 1} {{x - 1} \over {\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt x  + 1} \right)}}  \cr   &  = \mathop {\lim }\limits_{x \to 1} {1 \over {\left( {x + 1} \right)\left( {{{\root 3 \of x }^2} + \root 3 \of x  + 1} \right)}} - \mathop {\lim }\limits_{x \to 1} {1 \over {\left( {x + 1} \right)\left( {\sqrt x  + 1} \right)}}  \cr   &  = {1 \over {\left( {1 + 1} \right)\left( {1 + 1 + 1} \right)}} - {1 \over {\left( {1 + 1} \right)\left( {1 + 1} \right)}} = {1 \over 6} - {1 \over 4} =  - {1 \over {12}} \cr} \)

\( \Rightarrow \) Đáp án C đúng.

\(\eqalign{  & \mathop {\lim }\limits_{x \to 2} {{x - \sqrt {3x - 2} } \over {{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} {{{x^2} - 3x + 2} \over {\left( {{x^2} - 4} \right)\left( {x + \sqrt {3x - 2} } \right)}}  \cr   &  = \mathop {\lim }\limits_{x \to 2} {{\left( {x - 2} \right)\left( {x - 1} \right)} \over {\left( {x - 2} \right)\left( {x + 2} \right)\left( {x + \sqrt {3x - 2} } \right)}} = \mathop {\lim }\limits_{x \to 2} {{x - 1} \over {\left( {x + 2} \right)\left( {x + \sqrt {3x - 2} } \right)}} = {{2 - 1} \over {\left( {2 + 2} \right)\left( {2 + 2} \right)}} = {1 \over {16}} \cr} \)

\( \Rightarrow \) Đáp án D sai.

Chọn C.