Mọi người giải hộ em với ạ

Giải thích các bước giải:

$\begin{array}{l}
1)\lim \dfrac{{1 + 2 + 3 + ... + n}}{{{n^2} + 1}}\\
 = \lim \dfrac{{\dfrac{{\left( {1 + n} \right)n}}{2}}}{{{n^2} + 1}}\\
 = \lim \dfrac{{{n^2} + n}}{{2{n^2} + 2}}\\
 = \lim \dfrac{{1 + \dfrac{1}{n}}}{{2 + \dfrac{2}{{{n^2}}}}}\\
 = \dfrac{{1 + 0}}{{2 + 2.0}}\\
 = \dfrac{1}{2}\\
b)\lim \dfrac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{{n^2} + 2n + 7}}\\
 = \lim \dfrac{{\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}}{{{n^2} + 2n + 7}}\\
 = \lim \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6\left( {{n^2} + 2n + 7} \right)}}\\
 = \lim \dfrac{{2{n^3} + 3{n^2} + n}}{{6\left( {{n^2} + 2n + 7} \right)}}\\
 = \lim \dfrac{{2 + \dfrac{3}{n} + \dfrac{1}{{{n^2}}}}}{{6\left( {\dfrac{1}{n} + \dfrac{2}{{{n^2}}} + \dfrac{7}{{{n^3}}}} \right)}}\\
\lim \left( {2 + \dfrac{3}{n} + \dfrac{1}{{{n^2}}}} \right) = 2\\
\lim 6\left( {\dfrac{1}{n} + \dfrac{2}{{{n^2}}} + \dfrac{7}{{{n^3}}}} \right) = 0;\dfrac{1}{n} + \dfrac{2}{{{n^2}}} + \dfrac{7}{{{n^3}}} > 0\\
 \Rightarrow \lim \dfrac{{2 + \dfrac{3}{n} + \dfrac{1}{{{n^2}}}}}{{6\left( {\dfrac{1}{n} + \dfrac{2}{{{n^2}}} + \dfrac{7}{{{n^3}}}} \right)}} =  + \infty \\
 \Rightarrow \lim \dfrac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{{n^2} + 2n + 7}} =  + \infty \\
c)\lim \left[ {\left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right)} \right]\\
 = \lim \dfrac{{\left( {{2^2} - 1} \right)\left( {{3^2} - 1} \right)...\left( {{n^2} - 1} \right)}}{{{2^2}{3^2}...{n^2}}}\\
 = \lim \dfrac{{\left( {2 - 1} \right)\left( {2 + 1} \right)\left( {3 - 1} \right)\left( {3 + 1} \right)...\left( {n - 1} \right)\left( {n + 1} \right)}}{{{2^2}{3^2}...{n^2}}}\\
 = \lim \dfrac{{1.3.2.4.3.5.4.6....\left( {n - 1} \right)\left( {n + 1} \right)}}{{{2^2}{3^2}...{n^2}}}\\
 = \lim \dfrac{{{{1.2.3}^2}{{.4}^2}...{{\left( {n - 2} \right)}^2}{{\left( {n - 1} \right)}^2}n\left( {n + 1} \right)}}{{{2^2}{3^2}...{n^2}}}\\
 = \lim \dfrac{{n + 1}}{{2n}}\\
 = \lim \dfrac{{1 + \dfrac{1}{n}}}{2}\\
 = \dfrac{1}{2}\\
d)\lim \left( {\dfrac{1}{{1.3}} + \dfrac{1}{{3.5}} + ... + \dfrac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}} \right)\\
 = \lim \dfrac{1}{2}.\left( {\dfrac{2}{{1.3}} + \dfrac{2}{{3.5}} + ... + \dfrac{2}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}} \right)\\
 = \lim \dfrac{1}{2}\left( {1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + ... + \dfrac{1}{{2n - 1}} - \dfrac{1}{{2n + 1}}} \right)\\
 = \lim \dfrac{1}{2}\left( {1 - \dfrac{1}{{2n + 1}}} \right)\\
 = \dfrac{1}{2}\left( {1 - 0} \right)\\
 = \dfrac{1}{2}
\end{array}$