Giải thích các bước giải:
\(\begin{array}{l}
1.\mathop {\lim }\limits_{x \to \infty } \frac{{\left( {1 - \frac{3}{n}} \right)\left( {2 - \frac{1}{{{n^3}}}} \right)}}{{\frac{1}{n} + \frac{3}{{{n^4}}} - \frac{1}{{{n^5}}}}} = + \infty \\
3.\mathop {\lim }\limits_{x \to + \infty } \frac{{4 - \frac{2}{x} - \sqrt {1 + \frac{{12}}{x} + \frac{1}{{{x^2}}}} }}{{\sqrt {1 + \frac{{12}}{x}} + 5}} = \frac{{4 - 1}}{{1 + 5}} = \frac{1}{2}\\
4.\mathop {\lim }\limits_{x \to - \infty } \frac{{1 - \sqrt {4 + \frac{3}{x} + \frac{2}{{{x^2}}}} }}{{5 + \sqrt {1 + \frac{{11}}{{{x^2}}}} }} = \frac{{1 - 2}}{{5 + 1}} = - \frac{1}{6}\\
5.\mathop {\lim }\limits_{x \to - 1} \frac{{\left( {x + 1} \right)\left( {x + 6} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to - 1} \frac{{x + 6}}{{x - 1}} = \frac{{ - 1 + 6}}{{ - 1 - 1}} = - \frac{5}{2}\\
6.\mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {{x^2} + 6} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + 6}}{{x + 2}} = \frac{{10}}{4} = \frac{5}{2}\\
\end{array}\)
