Tính \(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + 3} - 3x + 1}}{{{x^2} - 1}}\).

* Hướng dẫn giải

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + 3}  - 3x + 1}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + 3}  - 2 - 3x + 3}}{{{x^2} - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + 3}  - 2}}{{{x^2} - 1}} - \mathop {\lim }\limits_{x \to 1} \dfrac{{3\left( {x - 1} \right)}}{{{x^2} - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\sqrt {{x^2} + 3}  - 2} \right)\left( {\sqrt {{x^2} + 3}  + 2} \right)}}{{\left( {{x^2} - 1} \right)\left( {\sqrt {{x^2} + 3}  + 2} \right)}} - \mathop {\lim }\limits_{x \to 1} \dfrac{3}{{x + 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 3 - 4}}{{\left( {{x^2} - 1} \right)\left( {\sqrt {{x^2} + 3}  + 2} \right)}} - \dfrac{3}{2}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\sqrt {{x^2} + 3}  + 2}} - \dfrac{3}{2} = \dfrac{1}{4} - \dfrac{3}{2} =  - \dfrac{5}{4}\end{array}\)

Chọn A