What is the reason that epsilon delta definition says the result of multiplying upper bound and $|x-a|$ is less than $\epsilon$?

Question

In epsilon delta definition when we want to prove a limit like $x^2$ at $x = 2$, we should find the upper bound of $|x+2|$ by assuming a random and small value for delta like $1$. next we say the upper bound times $|x-2|$ is less than $\epsilon$. How can we say this? we just know $|x-2||x+2| < \epsilon$ and $|x-2||x+2|<5|x-2|$. But how do we know $5|x-2| < \epsilon$? I hope I have asked my question correctly. Thanks.

Answer 1

We don't know that $5|x-2|<\varepsilon$. What we say here is: if we want to have $5|x-2|<\varepsilon$, that's the same thing as having $|x-2|<\frac\varepsilon5$ and then, in order to get that, we take $\delta=\frac\varepsilon5$, because then$$|x-2|<\delta\implies5|x-2|<\varepsilon.$$