Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\lim \frac{{\sqrt {{n^2} - 4n} - \sqrt {4{n^2} + 1} }}{{\sqrt {3{n^2} + 1} + n}}\\
= \lim \frac{{\frac{{{n^2} - 4n - 4{n^2} - 1}}{{\sqrt {{n^2} - 4n} + \sqrt {4{n^2} + 1} }}}}{{\sqrt {3{n^2} + 1} + n}}\\
= \lim \frac{{ - 3{n^2} - 4n - 1}}{{\left( {\sqrt {{n^2} + 4n} + \sqrt {4{n^2} + 1} } \right).\left( {\sqrt {3{n^2} + 1} + n} \right)}}\\
= \lim \frac{{\frac{{ - 3{n^2} - 4n - 1}}{{{n^2}}}}}{{\frac{{\sqrt {{n^2} + 4n} + \sqrt {4{n^2} + 1} }}{n}.\frac{{\sqrt {3{n^2} + 1} + n}}{n}}}\\
= \lim \frac{{ - 3 - \frac{4}{n} - \frac{1}{{{n^2}}}}}{{\left( {\sqrt {1 + \frac{4}{n}} + \sqrt {4 + \frac{1}{{{n^2}}}} } \right).\left( {\sqrt {3 + \frac{1}{{{n^2}}}} + 1} \right)}}\\
= \frac{{ - 3}}{{\left( {\sqrt 1 + \sqrt 4 } \right).\left( {\sqrt 3 + 1} \right)}}\\
= \frac{{ - 3}}{{3.\left( {\sqrt 3 + 1} \right)}} = - \frac{1}{{\sqrt 3 + 1}}\\
b,\\
\lim \frac{{\sqrt {4{n^2} + 1} - 2n - 1}}{{\sqrt {{n^2} + 4n + 1} - n}}\\
= \lim \left( {\frac{{4{n^2} + 1 - {{\left( {2n + 1} \right)}^2}}}{{\sqrt {4{n^2} + 1} + 2n + 1}}:\frac{{{n^2} + 4n + 1 - {n^2}}}{{\sqrt {{n^2} + 4n + 1} + n}}} \right)\\
= \lim \left( {\frac{{ - 4n}}{{\sqrt {4{n^2} + 1} + 2n + 1}}:\frac{{4n + 1}}{{\sqrt {{n^2} + 4n + 1} + n}}} \right)\\
= \lim \left( {\frac{{ - 4}}{{\sqrt {4 + \frac{1}{{{n^2}}}} + 2 + \frac{1}{n}}}:\frac{{4 + \frac{1}{n}}}{{\sqrt {1 + \frac{4}{n} + \frac{1}{{{n^2}}}} + 1}}} \right)\\
= \frac{{ - 4}}{{\sqrt 4 + 2}}:\frac{4}{{\sqrt 1 + 1}} = - \frac{1}{2}
\end{array}\)
