Proving $ \lim_{x\to 2} x^2 = 4$ I am trying to prove the following limit but I am a stuck at a step.

$$ \lim_{x\to 2} x^2 = 4$$

Proof : Let $\varepsilon &gt; 0$ be given. We want to find a $\delta &gt; 0$ such thant $0&lt;|x-2|&lt;\delta \rightarrow |x^2-4|&lt;\varepsilon$

$$ |x^2-4| &lt; \varepsilon$$
$$ |x-2||x+2| &lt; \varepsilon$$

What do I do from here ?
You could choose that $\;\delta&lt;\frac12\;$ , say, and then

$$|x-2|&lt;\delta\iff 2-\delta&lt;x&lt;2+\delta\implies \frac72=4-\delta&lt;x+2&lt;4+\delta=\frac92$$

so that you get

$$|x-2||x+2|&lt;|x-2|\cdot\frac92&lt;\frac{9\delta}2\color{red}{&lt;\epsilon}\iff \delta&lt;\frac{2\epsilon}9$$

and thus for any $\;\epsilon&gt;0\;$ , choosing $\;\delta=\min\left\{\frac12,\,\frac{2\epsilon}9\right\}\;$ you get what you want.

Observe that "the choosing" of $\;\delta\;$ that we began with is rather arbitrary. You could as well have chosen $\;\delta&lt;1\;,\;\;\delta&lt;0.001\;$ or whatever. We usually choose something more or less nice, though.
Hint: Fix $\epsilon&gt;0$.  Choose $\delta= \epsilon$, then we see that
\begin{align}
|x^2-4| = |x-2||x+2| \leq \epsilon |x+2|&lt; \epsilon(4+\epsilon)
\end{align}
whenever $|x-2|&lt;\delta$.