Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\lim \frac{{{{\left( {n + 1} \right)}^2}.{{\left( {2n - 1} \right)}^3}}}{{{n^5} + 3{n^2}}} = \lim \dfrac{{\frac{{{{\left( {n + 1} \right)}^2}}}{{{n^2}}}.\frac{{{{\left( {2n - 1} \right)}^3}}}{{{n^3}}}}}{{\frac{{{n^5} + 3{n^2}}}{{{n^5}}}}} = \lim \dfrac{{{{\left( {\frac{{n + 1}}{n}} \right)}^2}.{{\left( {\frac{{2n - 1}}{n}} \right)}^3}}}{{1 + \frac{3}{{{n^3}}}}} = \lim \dfrac{{{{\left( {1 + \frac{1}{n}} \right)}^2}.{{\left( {2 - \frac{1}{n}} \right)}^3}}}{{1 + \frac{3}{{{n^3}}}}} = \frac{{{{\left( {1 + 0} \right)}^2}.{{\left( {2 - 0} \right)}^3}}}{{1 + 0}} = 8\\
b,\\
\lim \frac{{{{3.4}^n} - {{2.4}^{n + 1}}}}{{{{5.4}^{n + 1}} + {4^n}}} = \lim \frac{{{4^n}\left( {3 - 2.4} \right)}}{{{4^n}.\left( {5.4 + 1} \right)}} = \lim \frac{{{4^n}.\left( { - 5} \right)}}{{{4^n}.21}} = \frac{{ - 5}}{{21}}\\
c,\\
\lim {q^n} = 0,\,\,\,\,\,\forall \left| q \right| < 1\\
\lim \frac{{{5^{n + 1}} - {3^{n - 1}}}}{{{5^{n + 2}} + {3^n}}} = \lim \dfrac{{\frac{{{5^{n + 1}} - {3^{n - 1}}}}{{{5^n}}}}}{{\frac{{{5^{n + 2}} + {3^n}}}{{{5^n}}}}} = \lim \dfrac{{5 - \frac{1}{3}.{{\left( {\frac{3}{5}} \right)}^n}}}{{{5^2} + {{\left( {\frac{3}{5}} \right)}^n}}} = \frac{{5 - \frac{1}{3}.0}}{{25 + 0}} = \frac{5}{{25}} = \frac{1}{5}\\
d,\\
\lim \frac{{\sqrt {{n^2} + n + 1}  + 3n}}{{2n - 3}} = \lim \dfrac{{n\sqrt {1 + \frac{1}{n} + \frac{1}{{{n^2}}}}  + 3n}}{{2n - 3}} = \lim \dfrac{{\sqrt {1 + \frac{1}{n} + \frac{1}{{{n^2}}}}  + 3}}{{2 - \frac{3}{n}}} = \frac{{\sqrt {1 + 0 + 0}  + 3}}{{2 - 0}} = 2
\end{array}\)