Đáp án:
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{1.\lim \frac{{\left[ {4{n^2} + 1 - {{\left( {2n - 1} \right)}^2}} \right]\left( {\sqrt {{n^2} + 2n} {\rm{ \;}} + n} \right)}}{{\left[ {{n^2} + 2n - {n^2}} \right]\left( {\sqrt {4{n^2} + 1} {\rm{ \;}} + 2n - 1} \right)}}}\\
{ = \lim \frac{{4n\left( {\sqrt {{n^2} + 2n} {\rm{ \;}} + n} \right)}}{{2n\left( {\sqrt {4{n^2} + 1} {\rm{ \;}} + 2n - 1} \right)}}}\\
{{\rm{\;}} = \lim \frac{{2\sqrt {{n^2} + 2n} {\rm{ \;}} + 2n}}{{\sqrt {4{n^2} + 1} {\rm{ \;}} + 2n - 1}}}\\
{ = \lim \frac{{2\sqrt {1 + \frac{2}{n}} {\rm{ \;}} + 2}}{{\sqrt {4 + \frac{1}{{{n^2}}}} {\rm{ \;}} + 2 - \frac{1}{n}}} = \frac{{2 + 2}}{{2 + 2}} = 1}\\
{2.\lim \left( {\frac{{\frac{1}{{{n^2}}}}}{{1 + \frac{1}{{{n^2}}}}} + \frac{{\frac{2}{{{n^2}}}}}{{1 + \frac{1}{{{n^2}}}}} + ... + \frac{{\frac{1}{n} - \frac{1}{{{n^2}}}}}{{1 + \frac{1}{{{n^2}}}}}} \right) = 0}\\
{}\\
{}
\end{array}\)
