The Problem:
Find a number $\delta$ such that if $|x-2| < \delta,$ then $|4x-8| < \epsilon$ where $\epsilon$$ = 1/10$
How I'd normally work it out: $$ \epsilon >|4x-8| $$ $$ \epsilon >4|x-2| $$ $$ \epsilon/4 >|x-2| $$
Let's take $\delta = \epsilon / 4$ and then go on to prove it.
Now my question is where does the $\epsilon=1/10$ part come in?
think about what the epsilon-delta proof is telling you, for every $\epsilon>|4x-8|$ you can find $\delta>|x-2|.$ now the $\frac{1}{10}$ is just an example of a value that you need to use the relationship you proved to get $\delta$, there is nothing special. it is just like when you first start to learn about functions sometimes you got question like "what is the value of $f(2)$ where $f(x)=x^2$"(ofc this example probably didn't happened to you but the general idea behind those 2 is the same, to understand what the relationship you showed means)