Giải thích các bước giải:
\(\begin{array}{l}
20,\\
\lim \frac{{2n}}{{{n^2} + 1}} = \lim \dfrac{{\frac{2}{n}}}{{1 + \frac{1}{{{n^2}}}}} = \frac{0}{{1 + 0}} = 0\\
21,\\
\lim \frac{{\sqrt {n + 1} }}{{n + 2}} = \lim \frac{{\sqrt {\frac{1}{n} + \frac{1}{{{n^2}}}} }}{{1 + \frac{2}{n}}} = \frac{0}{{1 + 0}} = 0\\
22,\\
\lim \frac{{2n + 3}}{{{n^2} + 1}} = \lim \dfrac{{\frac{2}{n} + \frac{3}{{{n^2}}}}}{{1 + \frac{1}{{{n^2}}}}} = \frac{{0 + 0}}{{1 + 0}} = 0\\
23,\\
\lim \frac{{{n^3} - 3{n^2} + 2}}{{{n^4} + 4{n^3} + 1}} = \lim \dfrac{{\frac{1}{n} - \frac{3}{{{n^2}}} + \frac{2}{{{n^4}}}}}{{1 + \frac{4}{n} + \frac{1}{{{n^4}}}}} = \frac{{0 - 0 + 0}}{{1 + 0 + 0}} = 0\\
24,\\
\lim \frac{{\sqrt[4]{{3{n^3} + 1}} - n}}{{\sqrt[4]{{2{n^4} + 3n + 1}} + n}} = \lim \left[ {\frac{{\sqrt[4]{{3{n^3} + 1}} - n}}{n}:\frac{{\sqrt[4]{{2{n^4} + 3n + 1}} + n}}{n}} \right]\\
= \lim \left[ {\left( {\sqrt[4]{{\frac{3}{n} + \frac{1}{{{n^4}}}}} - 1} \right):\left( {\sqrt[4]{{2 + \frac{3}{{{n^3}}} + \frac{1}{{{n^4}}}}} + 1} \right)} \right]\\
= \left( {0 - 1} \right):\left( {\sqrt[4]{2} + 1} \right)\\
= \frac{{ - 1}}{{\sqrt[4]{2} + 1}}
\end{array}\)
