Prove $\lim_{x\to 2} \sqrt{x^2+5} = 3$ As stated in the title, I need to prove that $\lim_{x\to 2} \sqrt{x^2+5} = 3$ using only the precise definition of a limit.

For any given $\varepsilon \gt 0$, there exists a $\delta = $
Such that $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} \rvert \lt \varepsilon $
I've attempted to convert $\sqrt{x^2+5}$ into the following:
$\sqrt{x^2+5} = \sqrt{(x^2-4)+9}\\
 \sqrt{x^2+5} = \sqrt{(x^2-4x+4)+4x +1} = \sqrt{(x-2)^2+4(x-2)+9} $
I have a hunch that I am heading in the wrong direction. Can I get some advice on what I might be doing wrong?

You are on the right track.
Also you can do this:
Let $\epsilon >0$
If $\delta<1$ then $1<x<3$ thus $|x+2|<5$

\begin{align}|\sqrt{x^2+5}-3|&=\frac{(\sqrt{x^2+5}-3)(\sqrt{x^2+5}+3)}{\sqrt{x^2+5}+3}\\&=\frac{|x^2-4|}{\sqrt{x^2+5}+3} \\&\leq \frac{|x+2||x-2|}{3}\\&\leq \frac{5}{3}|x-2|\end{align}

Now take $\delta \leq \min\{\frac{3}{5} \epsilon,1\}$ and you are done.
