$\displaystyle \displaystyle \begin{align*} \sqrt{n^2 + 3} + \sqrt{n^2 + 8} &> \sqrt{n^2} + \sqrt{n^2} \\ &= |n| + |n| \\ &= 2|n| \end{align*}$
Since $\displaystyle \displaystyle \begin{align*} 2|n| \to \infty \end{align*}$, therefore so must $\displaystyle \displaystyle \begin{align*} \sqrt{n^2 + 3} + \sqrt{n^2+8} \end{align*}$.
Quite frankly this is a trivial problem. As the others have said, as n goes to infinity, both parts do also and so their sum does.
A much more interesting problem would be $\displaystyle \lim_{x\to\infty}\left[\sqrt{n^2+ 3}- \sqrt{n^2+ 8}\right]$. For that, I would recommend multiplying by the "unit fraction" $\displaystyle \frac{\sqrt{n^2+ 3}+ \sqrt{n^2+ 8}}{\sqrt{n^2+3}+ \sqrt{n^2+ 8}}$.