Đáp án: -8/3
Giải thích các bước giải:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{\sqrt {x + 3} - \sqrt {2x - 1} - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {\sqrt {x + 3} - 2} \right) + \left( {1 - \sqrt {2x - 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\frac{{x + 3 - 4}}{{\sqrt {x + 3} + 2}} + \frac{{1 - 2x + 1}}{{1 + \sqrt {2x - 1} }}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right).\left( {\frac{1}{{\sqrt {x + 3} + 2}} - \frac{2}{{1 + \sqrt {2x - 1} }}} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x + 1}}{{\frac{1}{{\sqrt {x + 3} + 2}} - \frac{2}{{1 + \sqrt {2x - 1} }}}}\\
= \frac{{1 + 1}}{{\frac{1}{{\sqrt {1 + 3} + 2}} - \frac{2}{{1 + \sqrt {2 - 1} }}}}\\
= \frac{{ - 8}}{3}
\end{array}$
