Giải thích các bước giải:

\(\begin{array}{l}
a.\mathop {\lim }\limits_{x \to 1} \frac{{x + 1}}{{x - 2}} = \frac{{1 + 1}}{{1 - 2}} =  - 2\\
c.\mathop {\lim }\limits_{x \to  - \infty } {x^2}\left( {1 + \frac{1}{x} - \frac{1}{{{x^2}}}} \right) =  + \infty \\
Do:\mathop {\lim }\limits_{x \to  - \infty } {x^2} =  + \infty \\
\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \frac{1}{x} - \frac{1}{{{x^2}}}} \right) = 1\\
d.\mathop {\lim }\limits_{x \to  + \infty } \frac{{5 + \frac{2}{x} + \frac{3}{{{x^2}}}}}{{1 + \frac{1}{{{x^2}}}}} = 5\\
e.\mathop {\lim }\limits_{x \to  + \infty } \frac{{1 + \frac{7}{{{x^4}}}}}{{1 + \frac{1}{{{x^4}}}}} = 1\\
f.\mathop {\lim }\limits_{x \to  + \infty } \frac{{\frac{3}{{{x^3}}} - 1}}{{\frac{1}{x} + \frac{6}{{{x^4}}} + \frac{5}{{{x^5}}}}} =  - \infty \\
g.\mathop {\lim }\limits_{x \to  + \infty } \frac{{2 - \sqrt {3 + \frac{2}{{{x^2}}}} }}{{5 + \sqrt {1 + \frac{1}{{{x^2}}}} }} = \frac{{2 - \sqrt 3 }}{6}\\
h.\mathop {\lim }\limits_{x \to \infty } \frac{{2 - \frac{1}{{{x^2}}}}}{{\frac{3}{{{x^2}}} - 1}} =  - 2\\
i.\mathop {\lim }\limits_{x \to  - \infty } \frac{{ - 2 + \frac{1}{x} - \frac{3}{{{x^5}}}}}{{\frac{3}{{{x^3}}} - \frac{7}{{{x^5}}}}} =  + \infty \\
Do:\mathop {\lim }\limits_{x \to  - \infty } \frac{3}{{{x^3}}} - \frac{7}{{{x^5}}} =  - \infty \\
\mathop {\lim }\limits_{x \to  - \infty }  - 2 + \frac{1}{x} - \frac{3}{{{x^5}}} =  - 2\\
k.\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {4 + \frac{1}{{{x^2}}}}  - \sqrt {1 + \frac{5}{{{x^2}}}} }}{{2 - \frac{7}{x}}} = \frac{1}{2}\\
l.\mathop {\lim }\limits_{x \to  - \infty } \frac{{\frac{1}{x} + 3}}{{ - \sqrt {2 + \frac{3}{{{x^2}}}} }} =  - \frac{3}{{\sqrt 2 }}\\
m.\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} - x + 1 - {x^2}}}{{\sqrt {{x^2} - x + 1}  + x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{1 - x}}{{\sqrt {{x^2} - x + 1}  + x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\frac{1}{x} - 1}}{{\sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}}  + 1}} =  - \frac{1}{2}\\
n.\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {4{x^4} + {x^2}}  - {x^2}} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } {x^2}\left( { - \sqrt {4 + \frac{1}{{{x^2}}}}  - 1} \right) =  - \infty \\
o.\mathop {\lim }\limits_{x \to  - \infty } {x^5}\left( {4 - \frac{3}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}} \right) =  - \infty \\
Do:\mathop {\lim }\limits_{x \to  - \infty } {x^5} =  - \infty \\
\mathop {\lim }\limits_{x \to  - \infty } \left( {4 - \frac{3}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}} \right) = 4\\
p.\mathop {\lim }\limits_{x \to  + \infty } \frac{{8{x^3} + 2x - 8{x^3}}}{{\sqrt[3]{{{{\left( {8{x^3} + 2x} \right)}^2} + 2x.\sqrt[3]{{8{x^3} + 2x}} + 4{x^2}}}}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{2x}}{{\sqrt[3]{{{{\left( {8{x^3} + 2x} \right)}^2} + 2x.\sqrt[3]{{8{x^3} + 2x}} + 4{x^2}}}}}\\
 = 0
\end{array}\)