Giải thích các bước giải:
\(\begin{array}{l}
*)\\
lim\frac{{2 - 3n}}{{2n + 1}} = \lim \dfrac{{\frac{2}{n} - 3}}{{2 + \frac{1}{n}}} = \frac{{0 - 3}}{{2 + 0}} = - \frac{3}{2}\\
*)\\
\lim \frac{{{n^2} + 1}}{{{n^2} + 3}} = \lim \dfrac{{1 + \frac{1}{{{n^2}}}}}{{1 + \frac{3}{{{n^2}}}}} = \frac{{1 + 0}}{{1 + 0}} = 1\\
*)\\
\lim \frac{{\sqrt {{n^2} + 1} + 1}}{{3n + 4}} = \lim \dfrac{{\sqrt {{n^2}\left( {1 + \frac{1}{{{n^2}}}} \right)} + 1}}{{3n + 4}}\\
= \lim \dfrac{{n\sqrt {1 + \frac{1}{{{n^2}}}} + 1}}{{3n + 4}} = \lim \dfrac{{\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{1}{n}}}{{3 + \frac{4}{n}}} = \frac{{\sqrt 1 + 0}}{{3 + 0}} = \frac{1}{3}\\
*)\\
\lim \frac{{2n}}{{{n^2} + 1}} = \lim \dfrac{{\frac{2}{n}}}{{1 + \frac{1}{{{n^2}}}}} = \frac{0}{{1 + 0}} = 0
\end{array}\)
