Prove $f(x) = x^2 + 3$ is continuous at $x=3$ Prove that $f(x) = x^2 + 3$ is continuous at $x=3$.
I have tried using $\delta = \sqrt{\epsilon + 9} - 3$.
I tried to split $|x^2-9| = |(x-3)(x+3)|$ and tried to make $x+3$ in terms of $\delta$.
But I get $\delta^2 + 6\delta$.
I don't really get what I'm doing wrong or right so I need some help with finding the right $\delta$.
Also, I cannot really understand the goal of the proof, I read on elsewhere that I need to make $\delta$ as small as possible?

You need to solve the equation
$$0<|x-3|<\delta\implies|(x^2+3)-(3^2+3)|<\epsilon$$ for $\delta$.
As the function $x^2+3$ is monotonic around $x=3$, you can solve the equation
$$|x^2-9|=\epsilon$$ and use any $\delta$ such that $(3-\delta,3+\delta)$ is wholly contained between the two solutions, $x=\sqrt{9\pm\epsilon}.$ (Other possibilities are $x=-\sqrt{9\pm\epsilon}$, but this does not straddle $x=3$.)
The largest value of $\delta$ is thus
$$\min(\sqrt{9+\epsilon}-3,3-\sqrt{9-\epsilon})$$
and this is always $\sqrt{9+\epsilon}-3$.

Check:
$$-\delta<x-3<\delta\implies3-\delta<x<3+\delta\implies x^2<(6-\sqrt{9+\epsilon})^2,(\sqrt{9+\epsilon})^2.$$
But
$$(6-\sqrt{9+\epsilon})^2=36-12\sqrt{9+\epsilon}+9+\epsilon\ge 9-\epsilon$$
because
$$36+2\epsilon\ge 12\sqrt{9+\epsilon}$$
$$1296+144\epsilon+4\epsilon^2\ge 1296+144\epsilon$$
and $$\epsilon^2>0.$$
