Đáp án:
\[\lim \left( {\sqrt {{n^2} - 4an + 6} + \sqrt {9{n^2} + 6an + 3} - 4n} \right) = - a\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \left( {\sqrt {{n^2} - 4an + 6} + \sqrt {9{n^2} + 6an + 3} - 4n} \right)\\
= \lim \left[ {\left( {\sqrt {{n^2} - 4an + 6} - n} \right) + \left( {\sqrt {9{n^2} + 6an + 3} - 3n} \right)} \right]\\
= \lim \left[ {\frac{{{n^2} - 4an + 6 - {n^2}}}{{\sqrt {{n^2} - 4an + 6} + n}} + \frac{{9{n^2} + 6an + 3 - 9{n^2}}}{{\sqrt {9{n^2} + 6an + 3} + 3n}}} \right]\\
= \lim \left( {\frac{{ - 4an + 6}}{{\sqrt {{n^2} - 4an + 6} + n}} + \frac{{6an + 3}}{{\sqrt {9{n^2} + 6an + 3} + 3n}}} \right)\\
= \lim \left( {\frac{{ - 4a + \frac{6}{n}}}{{\sqrt {1 - \frac{{4a}}{n} + \frac{6}{{{n^2}}}} + 1}} + \frac{{6a + \frac{3}{n}}}{{\sqrt {9 + \frac{{6a}}{n} + \frac{3}{{{n^2}}}} + 3}}} \right)\\
= \frac{{ - 4a}}{{\sqrt 1 + 1}} + \frac{{6a}}{{\sqrt 9 + 3}}\\
= \frac{{ - 4a}}{2} + \frac{{6a}}{6} = - 2a + a = - a
\end{array}\)
