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