Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\lim \sqrt {{n^2} + 3n - 12} \\
= \lim \sqrt {{n^2}.\left( {1 + \frac{3}{n} - \frac{{12}}{{{n^2}}}} \right)} \\
= \lim n.\sqrt {1 + \frac{3}{n} - \frac{{12}}{{{n^2}}}} \\
= \left( { + \infty } \right).\sqrt 1 = + \infty \\
b,\\
\lim \sqrt {{n^4} + 12n - 10} \\
= \lim \sqrt {{n^4}\left( {1 + \frac{{12}}{{{n^3}}} - \frac{{10}}{{{n^4}}}} \right)} \\
= \lim \left[ {{n^2}.\sqrt {1 + \frac{{12}}{{{n^3}}} - \frac{{10}}{{{n^4}}}} } \right]\\
= \left( { + \infty } \right).\sqrt 1 = + \infty \\
c,\\
\lim \sqrt[3]{{{n^3} + 12{n^2} - 12}}\\
= \lim \sqrt[3]{{{n^3}.\left( {1 + \frac{{12}}{n} - \frac{{12}}{{{n^3}}}} \right)}}\\
= \lim \left[ {n.\sqrt[3]{{1 + \frac{{12}}{n} - \frac{{12}}{{{n^3}}}}}} \right]\\
= + \infty \\
d,\\
\lim \sqrt[3]{{27{n^3} + 12n - 7}}\\
= \lim \sqrt[3]{{{n^3}.\left( {27 + \frac{{12}}{{{n^2}}} - \frac{7}{{{n^3}}}} \right)}}\\
= \lim \left[ {n.\sqrt[3]{{27 + \frac{{12}}{{{n^2}}} - \frac{7}{{{n^3}}}}}} \right]\\
= \left( { + \infty } \right).3 = + \infty
\end{array}\)
