I don't understand the implication statement in a quadratic epsilon delta proof. If I have $\lim\limits_{x\to2} x^2=4$ and then  I use the $\epsilon-\delta$ proof, why is $|x+2| \leq 5$ , even though that is not possible because that would imply that $-5 \leq x+2 \leq 5$ but it is not that, it is $3 \leq x+ 2 \leq 5$? I meant to write less than not less than or equal to
In general, if you make some statement $P,$ and then you say "therefore $Q,$" you are not saying that $P$ and $Q$ are completely equivalent. There might be plenty of ways to satisfy $Q$ while $P$ is still false. What matters is that there is no way to make $P$ true and $Q$ false at the same time. In the examples you've given, we start with $P,$ a statement that is true for $x$ within some interval, and then make a statement $Q$ that is true for $x$ in a larger interval. No problem as long as the smaller interval fits completely inside the larger one.

Now, if someone were to say $3 &lt; x+2 &lt; 5,$ and then say that therefore
$\lvert x+2\rvert &lt; 4,$ then you would have a faulty "proof," because part of the interval that makes $3 &lt; x+2 &lt; 5$ also makes $\lvert x+2\rvert &lt; 4$ false, for example if $x=2.$
You claim that $\lim_{x\to 2}x^2=4$ with a $\epsilon-\delta$ argument. Fix $\varepsilon \in (0,1)$. Then it has to exist $\delta=\delta(\varepsilon)&gt;0$ such that
$$
|x-2|&lt;\delta \implies |x^2-4|&lt;\varepsilon,
$$
i.e.,
$$
2-\delta&lt;x&lt;2+\delta \implies 4-\varepsilon&lt;x^2&lt;4+\varepsilon.
$$
To this aim, fix $\delta=\varepsilon/5$ (which is $&lt;1$). Then
$$
x^2&lt;(2+\delta)^2=4+\delta^2+4\delta \le 4+5\delta = 4+\varepsilon
$$
and
$$
x^2&gt;(2-\delta)^2=4+\delta^2-4\delta&gt;4-4\delta&gt;4-5\delta=4-\varepsilon.
$$