Giải thích các bước giải:

\(\begin{array}{l}
a,\\
\lim \frac{{5{n^2} + 5n - 1}}{{\left( {5n + 2} \right)\left( {n - 4} \right)}} = \lim \frac{{\frac{{5{n^2} + 5n - 1}}{{{n^2}}}}}{{\frac{{5n + 2}}{n}.\frac{{n - 4}}{n}}} = \lim \frac{{1 + \frac{5}{n} - \frac{1}{{{n^2}}}}}{{\left( {5 + \frac{2}{n}} \right)\left( {1 - \frac{4}{n}} \right)}} = \frac{{1 + 0 - 0}}{{\left( {5 + 0} \right)\left( {1 - 0} \right)}} = \frac{1}{5}\\
b,\\
\lim \frac{{\sqrt {{n^2} + 4}  - \sqrt {4{n^2} + 1} }}{{\sqrt[3]{{27{n^3} - n}} + 3}} = \lim \frac{{n\sqrt {1 + \frac{4}{{{n^2}}}}  - n\sqrt {4 + \frac{1}{{{n^2}}}} }}{{n.\sqrt[3]{{27 - \frac{1}{{{n^2}}}}} + 3}} = \lim \frac{{\sqrt {1 + \frac{4}{{{n^2}}}}  - \sqrt {4 + \frac{1}{{{n^2}}}} }}{{\sqrt[3]{{27 - \frac{1}{{{n^2}}}}} + \frac{3}{n}}} = \frac{{\sqrt 1  - \sqrt 4 }}{{\sqrt[3]{{27}} + 0}} =  - \frac{1}{3}\\
c,\\
\lim {q^n} = 0,\,\,\,\forall \left| q \right| < 1\\
\lim \frac{{{5^{n + 1}} + {7^{n + 2}} + 1}}{{{3^{n + 1}} + {7^{n + 1}} + {{3.2}^n}}} = \lim \frac{{{{\left( {\frac{5}{7}} \right)}^{n + 1}} + 7 + {{\left( {\frac{1}{7}} \right)}^{n + 1}}}}{{{{\left( {\frac{3}{7}} \right)}^{n + 1}} + 1 + \frac{3}{2}.{{\left( {\frac{2}{7}} \right)}^{n + 1}}}} = \frac{7}{1} = 7\\
d,\\
\lim \left( {\sqrt {{n^2} + 1}  - \sqrt {{n^2} - 1} } \right)\\
 = \lim \frac{{\left( {\sqrt {{n^2} + 1}  - \sqrt {{n^2} - 1} } \right)\left( {\sqrt {{n^2} + 1}  + \sqrt {{n^2} - 1} } \right)}}{{\sqrt {{n^2} + 1}  + \sqrt {{n^2} - 1} }}\\
 = \lim \frac{{\left( {{n^2} + 1} \right) - \left( {{n^2} - 1} \right)}}{{\sqrt {{n^2} + 1}  + \sqrt {{n^2} - 1} }}\\
 = \lim \frac{2}{{\sqrt {{n^2} + 1}  + \sqrt {{n^2} - 1} }} = 0\\
\left( {\lim \left( {\sqrt {{n^2} + 1}  + \sqrt {{n^2} - 1} } \right) =  + \infty } \right)\\
e,\\
\lim \frac{{\sqrt {2{n^2} + 1}  - \sqrt {{n^2} + 1} }}{{n + 1}} = \lim \frac{{n\sqrt {2 + \frac{1}{{{n^2}}}}  - n\sqrt {1 + \frac{1}{{{n^2}}}} }}{{n + 1}}\\
 = \lim \frac{{\sqrt {2 + \frac{1}{{{n^2}}}}  - \sqrt {1 + \frac{1}{{{n^2}}}} }}{{1 + \frac{1}{n}}} = \frac{{\sqrt 2  - \sqrt 1 }}{1} = \sqrt 2  - 1\\
f,\\
\lim \sqrt {{n^4} + {n^2} - 2}  = \lim \left[ {{n^2}\sqrt {1 + \frac{1}{{{n^2}}} - \frac{2}{{{n^4}}}} } \right] =  + \infty 
\end{array}\)