Tính giới hạn sau \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x + 2} - 2x}}{{x - 1}}\).

* Hướng dẫn giải

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x + 2}  - 2x}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {2x + 2}  - 2x} \right)\left( {\sqrt {2x + 2}  + 2x} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {2x + 2}  + 2x} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{2x + 2 - 4{x^2}}}{{\left( {x - 1} \right)\left( {\sqrt {2x + 2}  + 2x} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{ - 2\left( {x - 1} \right)\left( {2x + 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {2x + 2}  + 2x} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{ - 2\left( {2x + 1} \right)}}{{\sqrt {2x + 2}  + 2x}}\\ = \frac{{ - 2.\left( {2.1 + 1} \right)}}{{\sqrt {2.1 + 2}  + 2.1}} =  - \frac{3}{2}\end{array}\)

Chọn D.