Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\lim \left( {{{3.2}^n} - {5^n}} \right)\\
= \lim \left[ {{5^n}.\left( {3.\frac{{{2^n}}}{{{5^n}}} - 1} \right)} \right]\\
= \lim \left[ {{5^n}.\left( {3.{{\left( {\frac{2}{5}} \right)}^n} - 1} \right)} \right]\\
= - \infty \\
\left( \begin{array}{l}
\lim {5^n} = + \infty \\
\lim {\left( {\frac{2}{5}} \right)^n} = 0 \Rightarrow \lim \left[ {3.{{\left( {\frac{2}{5}} \right)}^n} - 1} \right] = - 1
\end{array} \right)\\
b,\\
\lim \frac{{2{n^4} - {n^2} + n}}{{3{n^2} + 5n + 5}}\\
= \lim \dfrac{{2 - \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{\frac{3}{{{n^2}}} + \frac{5}{{{n^3}}} + \frac{5}{{{n^4}}}}}\\
= + \infty \\
\left( \begin{array}{l}
\lim \left( {2 - \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}}} \right) = 2\\
\lim \left( {\frac{3}{{{n^2}}} + \frac{5}{{{n^3}}} + \frac{5}{{{n^4}}}} \right) = {0^ + }
\end{array} \right)
\end{array}\)
