Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
a.\lim {n^3}\left( {\frac{3}{{{n^2}}} - 1} \right) = - \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } {n^3} = + \infty \\
\lim \left( {\frac{3}{{{n^2}}} - 1} \right) = - 1\\
d.\lim \left( {{3^n}} \right)\left( {\frac{{{3^4}{{.2.2}^n}}}{{{3^n}}} - 5} \right) = - \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } \left( {{3^n}} \right) = + \infty \\
\lim \left( {\frac{{{3^4}{{.2.2}^n}}}{{{3^n}}} - 5} \right) = - 5\\
C2:\\
a.\lim \frac{{4 + \frac{1}{n} + \frac{2}{{{n^2}}}}}{{a + \frac{5}{{{n^2}}}}} = \frac{4}{a} = 2\\
\to a = 2\\
b.\lim \frac{{{{\left( {2n + a} \right)}^3} - \left( {8{n^3} + 5} \right)}}{{{{\left( {2n + a} \right)}^2} + \sqrt[3]{{8{n^3} + 5}}.\left( {2n + a} \right) + \sqrt[3]{{{{\left( {8{n^3} + 5} \right)}^2}}}}}\\
= \lim \frac{{12{n^2}.a + 6n.{a^2} + {a^3} - 5}}{{{{\left( {2n + a} \right)}^2} + \sqrt[3]{{8{n^3} + 5}}.\left( {2n + a} \right) + \sqrt[3]{{{{\left( {8{n^3} + 5} \right)}^2}}}}}\\
= \lim \frac{{12a + \frac{{6{a^2}}}{n} + \frac{{{a^3}}}{{{n^2}}} - \frac{5}{{{n^2}}}}}{{{{\left( {2 + \frac{a}{n}} \right)}^2} + \sqrt[3]{{8 + \frac{5}{{{n^3}}}}}.\left( {2 + \frac{a}{n}} \right) + \sqrt[3]{{{{\left( {8 + \frac{5}{{{n^3}}}} \right)}^2}}}}}\\
= \frac{{12a}}{{4 + 2.2 + 4}} = \frac{{12.a}}{{12}} = 6\\
\to a = 6
\end{array}\)
