Giải thích các bước giải:
$\begin{array}{l}
g)\lim \dfrac{{{7^n} + {9^{n + 1}}}}{{{{2.9}^n} - {{3.5}^n}}}\\
= \lim \dfrac{{{{\left( {\dfrac{7}{9}} \right)}^n} + 9}}{{2 - 3.{{\left( {\dfrac{5}{9}} \right)}^n}}}\\
= \dfrac{9}{2}\\
h)\lim \dfrac{{{2^n}{{.3}^n} + {5^n}}}{{{6^{n - 1}} - 3}}\\
= \lim \dfrac{{1 + {{\left( {\dfrac{5}{6}} \right)}^n}}}{{\dfrac{1}{6} - 3.{{\left( {\dfrac{1}{6}} \right)}^n}}}\\
= \dfrac{1}{{\dfrac{1}{6}}}\\
= 6\\
i)\lim \dfrac{{{7^{2n}} - {9^n}}}{{{7^{2n - 1}} + {{3.2}^{2n}}}}\\
= \lim \dfrac{{{{49}^n} - {9^n}}}{{\dfrac{1}{7}{{.49}^n} + {{3.4}^n}}}\\
= \lim \dfrac{{1 - {{\left( {\dfrac{9}{{49}}} \right)}^n}}}{{\dfrac{1}{7} + 3.{{\left( {\dfrac{4}{{49}}} \right)}^n}}}\\
= \dfrac{1}{{\dfrac{1}{7}}}\\
= 7\\
k)\lim \dfrac{{{{\left( {{n^3} - 3} \right)}^3}{{\left( {3n + 1} \right)}^2}}}{{3n + 2{n^{11}}}}\\
= \lim \dfrac{{{{\left( {\dfrac{{{n^3} - 3}}{{{n^3}}}} \right)}^3}{{\left( {\dfrac{{3n + 1}}{n}} \right)}^2}}}{{3.\dfrac{1}{{{n^{10}}}} + 2}}\\
= \lim \dfrac{{{{\left( {1 - \dfrac{3}{{{n^3}}}} \right)}^3}{{\left( {3 + \dfrac{1}{n}} \right)}^2}}}{{3.\dfrac{1}{{{n^{10}}}} + 2}}\\
= \dfrac{{{1^3}{{.3}^2}}}{2}\\
= \dfrac{9}{2}
\end{array}$
$\begin{array}{l}
l)\lim \left( {\sqrt {{n^2} + n} - \sqrt[3]{{{n^3} + 2}}} \right)\\
= \lim \left( {\sqrt {{n^2} + n} - n + n - \sqrt[3]{{{n^3} + 2}}} \right)\\
= \lim \left( {\sqrt {{n^2} + n} - n} \right) + \lim \left( {n - \sqrt[3]{{{n^3} + 2}}} \right)\\
= \lim \dfrac{{{n^2} + n - {n^2}}}{{\sqrt {{n^2} + n} + n}} + \lim \dfrac{{{n^3} - \left( {{n^3} + 2} \right)}}{{{n^2} + n\sqrt[3]{{{n^3} + 2}} + {{\left( {\sqrt[3]{{{n^3} + 2}}} \right)}^2}}}\\
= \lim \dfrac{n}{{\sqrt {{n^2} + n} + n}} + \lim \dfrac{{ - 2}}{{{n^2} + n\sqrt[3]{{{n^3} + 2}} + {{\left( {\sqrt[3]{{{n^3} + 2}}} \right)}^2}}}\\
= \lim \dfrac{1}{{\sqrt {\dfrac{{{n^2} + n}}{{{n^2}}}} + 1}} + 0\\
= \lim \dfrac{1}{{\sqrt {1 + \dfrac{1}{n}} + 1}}\\
= \dfrac{1}{{\sqrt {1 + 0} + 1}}\\
= \dfrac{1}{2}
\end{array}$
