Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\lim \left( {3{n^3} - 7n - 6} \right)\\
= \lim \left[ {{n^3}.\left( {3 - \frac{7}{{{n^2}}} - \frac{6}{{{n^3}}}} \right)} \right]\\
= \lim {n^3}.\lim \left( {3 - \frac{7}{{{n^2}}} - \frac{6}{{{n^3}}}} \right)\\
= + \infty \\
\left( \begin{array}{l}
\lim {n^3} = + \infty \\
\lim \left( {3 - \frac{7}{{{n^2}}} - \frac{6}{{{n^3}}}} \right) = 3
\end{array} \right)\\
b,\\
\lim \left( { - 5{n^6} + 4{n^3} - 2n + 1} \right)\\
= \lim \left[ {{n^6}.\left( { - 5 + \frac{4}{{{n^3}}} - \frac{2}{{{n^5}}} + \frac{1}{{{n^6}}}} \right)} \right]\\
= \lim {n^6}.\lim \left( { - 5 + \frac{4}{{{n^3}}} - \frac{2}{{{n^5}}} + \frac{1}{{{n^6}}}} \right)\\
= - \infty \\
\left( \begin{array}{l}
\lim {n^6} = + \infty \\
\lim \left( { - 5 + \frac{4}{{{n^3}}} - \frac{2}{{{n^5}}} + \frac{1}{{{n^6}}}} \right) = - 5
\end{array} \right)\\
c,\\
\lim \left( {\sqrt {{n^2} - 4n} - n} \right)\\
= \lim \frac{{\left( {\sqrt {{n^2} - 4n} - n} \right)\left( {\sqrt {{n^2} - 4n} + n} \right)}}{{\sqrt {{n^2} - 4n} + n}}\\
= \lim \frac{{\left( {{n^2} - 4n} \right) - {n^2}}}{{\sqrt {{n^2} - 4n} + n}}\\
= \lim \frac{{ - 4n}}{{\sqrt {{n^2} - 4n} + n}}\\
= \lim \frac{{ - 4}}{{\sqrt {1 - \frac{4}{n}} + 1}}\\
= \frac{{ - 4}}{{\sqrt 1 + 1}}\\
= - 2
\end{array}\)
