Đáp án:
\[\lim \left( {2n - \sqrt {9{n^2} + n} + \sqrt {{n^2} + 2n} } \right) = \frac{5}{6}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \left( {2n - \sqrt {9{n^2} + n} + \sqrt {{n^2} + 2n} } \right)\\
= \lim \left[ {\left( {3n - \sqrt {9{n^2} + n} } \right) + \left( {\sqrt {{n^2} + 2n} - n} \right)} \right]\\
= \lim \left[ {\frac{{\left( {3n - \sqrt {9{n^2} + n} } \right)\left( {3n + \sqrt {9{n^2} + n} } \right)}}{{3n + \sqrt {9{n^2} + n} }} + \frac{{\left( {\sqrt {{n^2} + 2n} - n} \right)\left( {\sqrt {{n^2} + 2n} + n} \right)}}{{\sqrt {{n^2} + 2n} + n}}} \right]\\
= \lim \left[ {\frac{{{{\left( {3n} \right)}^2} - \left( {9{n^2} + n} \right)}}{{3n + \sqrt {9{n^2} + n} }} + \frac{{\left( {{n^2} + 2n} \right) - {n^2}}}{{\sqrt {{n^2} + 2n} + n}}} \right]\\
= \lim \left[ {\frac{{ - n}}{{3n + \sqrt {9{n^2} + n} }} + \frac{{2n}}{{\sqrt {{n^2} + 2n} + n}}} \right]\\
= \lim \left[ {\frac{{ - 1}}{{3 + \sqrt {9 + \frac{1}{n}} }} + \frac{2}{{\sqrt {1 + \frac{2}{n}} + 1}}} \right]\\
= \frac{{ - 1}}{{3 + \sqrt {9 + 0} }} + \frac{2}{{\sqrt {1 + 0} + 1}}\\
= \frac{{ - 1}}{6} + 1\\
= \frac{5}{6}
\end{array}\)
