Đáp án: -1/3
Giải thích các bước giải:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to - 2} \frac{{x - 1 + \sqrt {2{x^2} + 1} }}{{x + 2}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{{{\left( {x - 1} \right)}^2} - \left( {2{x^2} + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1 - \sqrt {2{x^2} + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - 2x + 1 - 2{x^2} - 1}}{{\left( {x + 2} \right)\left( {x - 1 - \sqrt {2{x^2} + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{ - {x^2} - 2x}}{{\left( {x + 2} \right)\left( {x - 1 - \sqrt {2{x^2} + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{ - x\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 1 - \sqrt {2{x^2} + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{ - x}}{{\left( {x - 1 - \sqrt {2{x^2} + 1} } \right)}}\\
= \frac{2}{{\left( { - 3 - 3} \right)}} = \frac{{ - 1}}{3}
\end{array}$
