<s>[INST] 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?
 [/INST] You're doing fine! So, in order to have$$\left|\sqrt{(x-2)^2+4(x-2)+9}-3\right|<\varepsilon,$$all you need is to have\begin{multline}\left|\sqrt{(x-2)^2+4(x-2)+9}-3\right|.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|<\\<\varepsilon.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|,\end{multline}or, in other words,$$\bigl|(x-2)^2+4(x-2)\bigr|<\varepsilon.\left|\sqrt{(x-2)^2+4(x-2)+9}+3\right|.$$In order for this to happen, all you need is that $\bigl|(x-2)^2+4(x-2)\bigr|<3\varepsilon$. So, pick $\delta$ such that $\delta\leqslant\frac38\varepsilon$ and also $\delta\leqslant\sqrt{\frac32\varepsilon}$. Then, if $|x-2|<\delta$,$$\bigl|(x-2)^2+4(x-2)\bigr|\leqslant|x-2|^2+4|x-2|\leqslant\frac32\varepsilon+\frac32\varepsilon=3\varepsilon.$$
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