Giải thích các bước giải:
\(\begin{array}{l}
a.\lim n\sqrt {1 + \frac{3}{n} + \frac{1}{{{n^2}}}} = + \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } n = + \infty \\
\lim \sqrt {1 + \frac{3}{n} + \frac{1}{{{n^2}}}} = 1\\
b.\lim n\left( {\sqrt[3]{{ - 8 + \frac{2}{n} - \frac{3}{{{n^3}}}}}} \right) = - \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } n = + \infty \\
\lim \left( {\sqrt[3]{{ - 8 + \frac{2}{n} - \frac{3}{{{n^3}}}}}} \right) = - 2\\
c.\lim \frac{{{n^2} + 3 - 4{n^2} + 1}}{{\sqrt {{n^2} + 3} - \sqrt {4{n^2} + 1} }}\\
= \lim \frac{{ - 3{n^2} + 4}}{{\sqrt {{n^2} + 3} - \sqrt {4{n^2} + 1} }}\\
= \lim \frac{{ - 3 + \frac{4}{{{n^2}}}}}{{\sqrt {1 + \frac{3}{{{n^2}}}} - \sqrt {4 + \frac{1}{{{n^2}}}} }} = \frac{{ - 3}}{{1 - 2}} = 3\\
d.\lim \frac{{\sqrt {4 - \frac{1}{{{n^2}}}} }}{{\frac{1}{n} - 3}} = \frac{{ - 2}}{3}\\
e.\lim \frac{{{n^2} + 2n - \left( {{n^2} + 2n + 1} \right)}}{{\sqrt {{n^2} + 2n} + n + 1}}\\
= \lim \frac{{ - 1}}{{\sqrt {{n^2} + 2n} + n + 1}} = 0\\
f.\lim \frac{{3n - 1 - 2n + 1}}{{\sqrt {3n - 1} + \sqrt {2n - 1} }}\\
= \lim \frac{n}{{\sqrt {3n - 1} + \sqrt {2n - 1} }}\\
= \lim \frac{1}{{\sqrt {\frac{3}{n} - \frac{1}{{{n^2}}}} + \sqrt {\frac{2}{n} - \frac{1}{{{n^2}}}} }} = + \infty
\end{array}\)
