Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
*)\\
\mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} - 1}}{{{x^2} + 5x + 4}} = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 4} \right)}} = \mathop {\lim }\limits_{x \to  - 1} \frac{{x - 1}}{{x + 4}} = \frac{{ - 1 - 1}}{{ - 1 + 4}} = \frac{{ - 2}}{3}\\
*)\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x + 7}  - 3}}{{{x^2} - 4x + 3}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {2x + 7}  - 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}{{\left( {{x^2} - 4x + 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {2x + 7} \right) - {3^2}}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{2\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{2}{{\left( {x - 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \frac{2}{{\left( {1 - 3} \right)\left( {\sqrt {2.1 + 7}  + 3} \right)}}\\
 =  - \frac{1}{6}\\
*)\\
\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 2x - 8}}{{{x^2} - 3x + 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x + 4} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x + 4}}{{x - 1}} = \frac{{2 + 4}}{{2 - 1}} = 6\\
*)\\
\mathop {\lim }\limits_{x \to \infty } \frac{{{x^4} - 2x}}{{\left( {{x^2} + 1} \right)\left( {2{x^2} - 5} \right)}}\\
 = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{x^4} - 2x}}{{{x^4}}}}}{{\frac{{{x^2} + 1}}{{{x^2}}}.\frac{{2{x^2} - 5}}{{{x^2}}}}}\\
 = \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \frac{2}{{{x^3}}}}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)\left( {2 - \frac{5}{{{x^2}}}} \right)}}\\
 = \frac{{1 - 0}}{{\left( {1 + 0} \right)\left( {2 - 0} \right)}} = \frac{1}{2}\\
*)\\
\mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{{x - 1}} - \frac{2}{{{x^2} - 1}}} \right)\\
 = \mathop {\lim }\limits_{x \to 1} \left( {\frac{1}{{x - 1}} - \frac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right)\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 1} \right) - 2}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{1}{{x + 1}} = \frac{1}{{1 + 1}} = \frac{1}{2}
\end{array}\)