Show that $f(x)=x^2$ is continuous at $a=2$ using the $\delta-\epsilon$ definition of continuity.

So we want to find a $\delta>0$
such that for all $2-\delta<x<2+\delta$
, we will have $4-\epsilon<x^{2}<4+\epsilon$ for all $\epsilon>0$
. If we can find a way to express $\delta$ as a function $\delta (\epsilon)$, $\delta:\mathbb{R}_+ \to \mathbb{R}_+ $ then we will have solved the problem. But I can’t see how to relate $\epsilon$ to $\delta$ in this case.

My initial reaction is write $(2-\delta)^2<x^2<(2+\delta)^2$, this makes the two inequalities look related, but I am not sure where to go from here.

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I think the way you write the inequalities obscures the point of view. The canonical way of looking at it, is that you want $\delta>0$ such that $|x-2|<\delta$ implies $|x^2-4|<\varepsilon$.

Fix $\varepsilon>0$. You have
$$
|x^2-4|=|x+2|\,|x-2|<\delta\,|x+2|<\delta(2+\delta+2)=\delta(4+\delta).
$$
So you need to choose $\delta>0$ such that $\delta(4+\delta)<\varepsilon$. As the function $t\mapsto t(t+4)$ is increasing for $t>0$, we look at $t(t+4)=\varepsilon$, i.e. $t^2+4t-\varepsilon=0$. The positive $t$ satisfying this is
$$
\frac{-4+\sqrt{16+4\varepsilon}}2=-2+\sqrt{4+\varepsilon}.
$$
So any $\delta<-2+\sqrt{4+\varepsilon}$ will do. For instance you can take $-2+\sqrt{4+\varepsilon/2}$.

Another approach would be to obtain two estimates one for $|x-2|$ and another for $|x + 2|$.

For instance, we want to find $\delta$ such that if $|x – 2 | < \delta \implies |x^2 – 4 | < \epsilon$.

First, suppose $|x – 2 | < 1$, and now using the triangle inequality

$$ | x + 2 | = |x + 4 – 2| \leq |x -2| + 4 < 5 $$

$$ \therefore |x^2 – 4| = |x-2||x+2| < 5 \delta. $$

To make the above estimate less that your $\epsilon$, it should be obvious that we msut choose $ \delta’ = \frac{\epsilon}{5}$. But we already took $\delta = 1$ to obtain the bound for $|x + 2 |$. Therefore, $\delta = \min\{1, \delta’\}$.