Đáp án:
\[\lim \frac{{\sqrt {4{n^2} + 3} - 2n + 1}}{{n.\left( {\sqrt {{n^2} + 3} - 2n} \right)}} = 0\]
Giải thích các bước giải:
\(\begin{array}{l}
\lim \frac{{\sqrt {4{n^2} + 3} - 2n + 1}}{{n.\left( {\sqrt {{n^2} + 3} - 2n} \right)}}\\
= \lim \frac{{\frac{{4{n^2} + 3 - {{\left( {2n - 1} \right)}^2}}}{{\sqrt {4{n^2} + 3} + 2n - 1}}}}{{n.\frac{{{n^2} + 3 - 4{n^2}}}{{\sqrt {{n^2} + 3} + 2n}}}}\\
= \lim \frac{{\frac{{4n + 2}}{{\sqrt {4{n^2} + 3} + 2n - 1}}}}{{n.\frac{{ - 3{n^2} + 3}}{{\sqrt {{n^2} + 3} + 2n}}}}\\
= \lim \left( {\frac{{4n + 2}}{{\sqrt {4{n^2} + 3} + 2n - 1}}:\frac{{ - 3{n^3} + 3n}}{{\sqrt {{n^2} + 3} + 2n}}} \right)\\
= \lim \left( {\frac{{4 + \frac{2}{n}}}{{\sqrt {4 + \frac{3}{{{n^2}}}} + 2 - \frac{1}{n}}}:\frac{{ - 3{n^2} + 3}}{{\sqrt {1 + \frac{3}{{{n^2}}}} + 2}}} \right)\\
\lim \frac{{4 + \frac{2}{n}}}{{\sqrt {4 + \frac{3}{{{n^2}}}} + 2 - \frac{1}{n}}} = \frac{4}{{\sqrt 4 + 2}} = 1\\
\lim \frac{{ - 3{n^2} + 3}}{{\sqrt {1 + \frac{3}{{{n^2}}}} + 2}} = - \infty \\
\Rightarrow \lim \left( {\frac{{4 + \frac{2}{n}}}{{\sqrt {4 + \frac{3}{{{n^2}}}} + 2 - \frac{1}{n}}}:\frac{{ - 3{n^2} + 3}}{{\sqrt {1 + \frac{3}{{{n^2}}}} + 2}}} \right) = \frac{1}{{ - \infty }} = 0
\end{array}\)
