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. :-)
 A: An alternative to "rationalising the denominator" is
$$
\begin{align}
-3n+\sqrt{9n^2+11n+16}
&=-3n+3n\sqrt{1+\frac{11}{9n}+\frac{16}{9n^2}}\\
&=-3n+3n\left(1+\frac{11}{9n}+O(n^{-2})\right)^{1/2}\\
&=-3n+3n\left(1+\frac{11}{18n}+O(n^{-2})\right)\\
&=\frac{11}{6}+O(n^{-1}).
\end{align}\
$$
A: As hinted in the comments,
$$\lim_{n \rightarrow \infty}{a_n}=\lim_{n \rightarrow \infty}{\bigg(-3n+\sqrt{{9n^2}+11n+16}\bigg)}=\lim_{n \rightarrow \infty}{\bigg(\bigg(-3n+\sqrt{{9n^2}+11n+16}\bigg)\cdot\bigg(\frac{-3n-\sqrt{{9n^2}+11n+16}}{-3n-\sqrt{{9n^2}+11n+16}}\bigg)\bigg)}=\lim_{n \rightarrow \infty}\bigg(\frac{9n^2 - ({9n^2}+11n+16)}{-3n-\sqrt{{9n^2}+11n+16}}\bigg)=\lim_{n \rightarrow \infty}\bigg(\frac{-11n - 16}{-3n-\sqrt{{9n^2}+11n+16}}\bigg)=\lim_{n \rightarrow \infty}\bigg(\frac{11n + 16}{3n+\sqrt{{9n^2}+11n+16}}\bigg)=\lim_{n \rightarrow \infty}\bigg(\frac{11 + (16/n)}{3+\sqrt{9+(11/n)+(16/n^2)}}\bigg)=\frac{11+0}{3+\sqrt{9+0+0}}=\frac{11}{3+3}=\frac{11}{6}.$$
A: \begin{align*}
 -3n + \sqrt{9n^2+11n+16}&=\frac{(-3n + \sqrt{9n^2+11n+16})(-3n - \sqrt{9n^2+11n+16})}{-3n - \sqrt{9n^2+11n+16}}\\
&=\frac{9n^2-(9n^2+11n+16)}{-3n - \sqrt{9n^2+11n+16}}
\end{align*}
