Ok say we were to prove this simple limit: $\lim \limits_{x \to 2}$ $x^2$
For all epsilon > 0, there exist a delta > 0 such that IF $ 0 < |x-2|< \delta$ THEN $|x^2-4| < \epsilon$
We know that $|x^2-4|$ = $|x+2||x-2| < \epsilon$
Now we need to find an estimated value for $|x+2|$
Let $\delta \le 1$
$|x-2| < \delta \le 1$
So, $|x-2| < 1$
Which tells us, $1 \lt x \lt 3$
Add 2 to all sides, $3 \lt x+2 \lt 5$
The example I had looked at takes $|x+2| < 5$, Why can you not take $|x+2| \gt 3$
This step is what gets me confused, I do understand why the first case is taken rather than the second one , and I do not know how to check if the one I had taken works.
Of course you can use the lower bound; but it is not useful here. As the definition shows, our goal is to see, given any $\varepsilon > 0$, whether we can make $|x^{2}-4| < \varepsilon$ by bounding $|x-2|$ from above in terms of $\varepsilon$; this indicates that it is best for us to bound $|x+2|$ from above. However, it is not necessary to bound $|x+2|$ in the way you are considering.
If $x \in \mathbb{R}$, then $|x^{2}-4| = |x-2||x+2|$; to bound $|x+2|$ from above, we find that for $|x-2| < 1$ we have $ |x| - 2 \leq |x-2| < 1, $ which implies that $|x+2| \leq |x| + 2 < 5$, and hence $|x-2||x+2| < 5|x-2|$; if $\varepsilon > 0$, then $|x-2| < \varepsilon/5$ only if $5|x-2| < \varepsilon$. Combining the two upper bounds for $|x-2|$ together, we have obtained this: for every $\varepsilon > 0$, we have $|x-2| < \min \{ 1, \varepsilon/5 \}$ only if $|x^{2} - 4| < \varepsilon$.
I hope the above somewhat helps.