Đáp án:
\(\lim \dfrac{{\sqrt {4{n^2} + 1} - 2n - 1}}{{\sqrt {{n^2} + 4n + 1} - n}}= - \dfrac{1}{4}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \dfrac{{\sqrt {4{n^2} + 1} - 2n - 1}}{{\sqrt {{n^2} + 4n + 1} - n}}\\
= \lim \left( {\dfrac{{{{\sqrt {4{n^2} + 1} }^2} - {{\left( {2n + 1} \right)}^2}}}{{\sqrt {4{n^2} + 1} + 2n + 1}}:\dfrac{{{{\sqrt {{n^2} + 4n + 1} }^2} - {n^2}}}{{\sqrt {{n^2} + 4n + 1} + n}}} \right)\\
= \lim \left( {\dfrac{{4{n^2} + 1 - 4{n^2} - 4n - 1}}{{\sqrt {4{n^2} + 1} + 2n + 1}}:\dfrac{{{n^2} + 4n + 1 - {n^2}}}{{\sqrt {{n^2} + 4n + 1} + n}}} \right)\\
= \lim \left( {\dfrac{{ - 4n}}{{\sqrt {4{n^2} + 1} + 2n + 1}}:\dfrac{{4n + 1}}{{\sqrt {{n^2} + 4n + 1} + n}}} \right)\\
= \lim \left( {\dfrac{{ - 4}}{{\sqrt {4 + \frac{1}{{{n^2}}}} + 2 + \dfrac{1}{n}}}:\dfrac{{4 + \frac{1}{n}}}{{\sqrt {1 + \dfrac{4}{n} + \dfrac{1}{{{n^2}}}} + 1}}} \right)\\
= \dfrac{{ - 2}}{{\sqrt 4 + 2}}:\dfrac{4}{{\sqrt 1 + 1}} = \dfrac{{ - 2}}{4}:\dfrac{4}{2} = - \dfrac{1}{4}
\end{array}\)
