What's the trick to get the limit of a sequence with roots like $a_n = -3n + \sqrt{9n^2+11n+16}$? I know of some "tricks" for limits of sequences like
$$b_n = \frac{n^2+7}{3n+1} = \frac{n^2+7}{3n+1} \cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}} = \frac{\frac{n^2}{n^2}+\frac{7}{n^2}}{\frac{3n}{n^2}+\frac{1}{n^2}}$$
Surely, there has to be a trick like this for the above example $a_n = -3n + \sqrt{9n^2+11n+16}$? Thanks. :-)

An alternative to "rationalising the denominator" is
$$
\begin{align}
-3n+\sqrt{9n^2+11n+16}
&=-3n+3n\sqrt{1+\frac{11}{9n}+\frac{16}{9n^2}}\\
&=-3n+3n\left(1+\frac{11}{9n}+O(n^{-2})\right)^{1/2}\\
&=-3n+3n\left(1+\frac{11}{18n}+O(n^{-2})\right)\\
&=\frac{11}{6}+O(n^{-1}).
\end{align}\
$$
