Giải thích các bước giải:

\(\begin{array}{l}
1,\\
\frac{1}{{\left( {2n - 1} \right).\left( {2n + 1} \right)}} = \frac{1}{2}.\frac{2}{{\left( {2n - 1} \right).\left( {2n + 1} \right)}} = \frac{1}{2}.\frac{{\left( {2n + 1} \right) - \left( {2n - 1} \right)}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \frac{1}{2}\left[ {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right]\\
 \Rightarrow {u_n} = \frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + .... + \frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}\\
 = \frac{1}{2}.\left( {1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + ..... + \frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)\\
 = \frac{1}{2}.\left( {1 - \frac{1}{{2n + 1}}} \right)\\
 \Rightarrow \lim {u_n} = \lim \left( {\frac{1}{2} - \frac{1}{{2.\left( {2n + 1} \right)}}} \right) = \frac{1}{2}\\
2,\\
\lim \frac{{2{n^2} - 3}}{{ - 2{n^3} - 4}} = \lim \frac{{2 - \frac{3}{{{n^2}}}}}{{ - 2n - \frac{4}{{{n^2}}}}} = \lim \frac{{ - 1}}{n} = 0\\
\lim \frac{{2{n^2} - 3}}{{ - 2{n^2} - 1}} = \lim \frac{{2 - \frac{3}{{{n^2}}}}}{{ - 2 - \frac{1}{{{n^2}}}}} = \frac{2}{{ - 2}} =  - 1\\
\lim \frac{{2{n^2} - 3}}{{ - 2{n^3} - 2{n^2}}} = \lim \frac{{2 - \frac{3}{{{n^2}}}}}{{ - 2n - 2}} = \lim \frac{{ - 1}}{{n + 1}} = 0\\
\lim \frac{{2{n^3} - 3}}{{ - 2{n^2} - 1}} = \lim \frac{{2n - \frac{3}{{{n^2}}}}}{{ - 2 - \frac{1}{{{n^2}}}}} = \lim \left( { - n} \right) =  - \infty \\
3,\\
\lim \frac{{{n^3} + 4n - 5}}{{3{n^3} + {n^2} + 7}} = \lim \frac{{1 + \frac{4}{{{n^2}}} - \frac{5}{{{n^3}}}}}{{3 + \frac{1}{n} + \frac{7}{{{n^3}}}}} = \frac{1}{3}
\end{array}\)