Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
16,\\
\lim \frac{{4n + 5}}{{n - 1}} = \lim \frac{{4 + \frac{5}{n}}}{{1 - \frac{1}{n}}} = \frac{{4 + 0}}{{1 - 0}} = 4\\
17,\\
a,\\
\mathop {\lim }\limits_{x \to  - 2} \frac{{2x + 3}}{{x + 1}} = \frac{{2.\left( { - 2} \right) + 3}}{{ - 2 + 1}} = 1\\
b,\\
\mathop {\lim }\limits_{x \to  - \infty } \left( {2{x^2} + 3x - 2} \right) = \mathop {\lim }\limits_{x \to  - \infty } \left[ {{x^2}\left( {2 + \frac{3}{x} - \frac{2}{{{x^2}}}} \right)} \right]\\
\left. \begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \left( {{x^2}} \right) =  + \infty \\
\mathop {\lim }\limits_{x \to  - \infty } \left( {2 + \frac{3}{x} - \frac{2}{{{x^2}}}} \right) = 2
\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to  - \infty } \left[ {{x^2}\left( {2 + \frac{3}{x} - \frac{2}{{{x^2}}}} \right)} \right] =  + \infty \\
c,\\
\mathop {\lim }\limits_{x \to  - 2} \frac{{{x^2} - 4}}{{x + 2}} = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{x + 2}} = \mathop {\lim }\limits_{x \to  - 2} \left( {x - 2} \right) =  - 2 - 2 =  - 4\\
d,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3x + 1}  - 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {3x + 1}  - 2} \right)\left( {\sqrt {3x + 1}  + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1}  + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {3x + 1} \right) - {2^2}}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{3\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1}  + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{3}{{\sqrt {3x + 1}  + 2}} = \frac{3}{{\sqrt {3.1 + 1}  + 2}} = \frac{3}{4}\\
e,\\
\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} + 1}  - x + 1}}{{x + 3}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)}  - x + 1}}{{x + 3}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|.\sqrt {1 + \frac{1}{{{x^2}}}}  - x + 1}}{{x + 3}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {1 + \frac{1}{{{x^2}}}}  - x + 1}}{{x + 3}}\,\,\,\,\,\,\,\left( {x \to  - \infty  \Rightarrow x < 0 \Rightarrow \left| x \right| =  - x} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {1 + \frac{1}{{{x^2}}}}  - 1 + \frac{1}{x}}}{{1 + \frac{3}{x}}} = \frac{{ - \sqrt 1  - 1}}{1} =  - 2\\
f,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3x + 1}  + \sqrt[3]{{2x - 1}} - 3}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\sqrt {3x + 1}  - 2}}{{x - 1}} + \frac{{\sqrt[3]{{2x - 1}} - 1}}{{x - 1}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\left( {\sqrt {3x + 1}  - 2} \right)\left( {\sqrt {3x + 1}  + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1}  + 2} \right)}} + \frac{{\left( {\sqrt[3]{{2x - 1}} - 1} \right)\left( {{{\sqrt[3]{{2x - 1}}}^2} + \sqrt[3]{{2x - 1}}.1 + {1^2}} \right)}}{{\left( {x - 1} \right)\left( {{{\sqrt[3]{{2x - 1}}}^2} + \sqrt[3]{{2x - 1}}.1 + {1^2}} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\left( {3x + 1} \right) - {2^2}}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1}  + 2} \right)}} + \frac{{\left( {2x - 1} \right) - {1^3}}}{{\left( {x - 1} \right)\left( {{{\sqrt[3]{{2x - 1}}}^2} + \sqrt[3]{{2x - 1}} + 1} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{3\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1}  + 2} \right)}} + \frac{{2\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{{\sqrt[3]{{2x - 1}}}^2} + \sqrt[3]{{2x - 1}} + 1} \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{3}{{\sqrt {3x + 1}  + 2}} + \frac{2}{{{{\sqrt[3]{{2x - 1}}}^2} + \sqrt[3]{{2x - 1}} + 1}}} \right]\\
 = \frac{3}{{\sqrt {3.1 + 1}  + 2}} + \frac{2}{{{{\sqrt[3]{{2.1 - 1}}}^2} + \sqrt[3]{{2.1 - 1}} + 1}}\\
 = \frac{3}{4} + \frac{2}{3} = \frac{{17}}{{12}}
\end{array}\)