# proving limits with Epsilon-Delta definition

Say I've got $\lim_{x\to a} f(x)= A$ and $\lim_{x\to a} g(x)= B$

How do I do the following:

1. prove that if A < B, then there is the existence of a $\delta$ such that when $0 < |x-a| < \delta$, we have $f(x) < g(x)$;

2. Find an example that there exists a $\delta$ such that when $0 < |x-a| < \delta$ , we have $f(x) < g(x)$ but $A = B$

I have a rough solution for part (i), but I'm not sure if I'm right so I'm just going to put it out here: from $A < B$, we have $0 < |x-a| < \delta$ implies that $f(x)<g(x)$

From here, we have $\lim_{x\to a} f(x) < \lim_{x\to a} g(x)$

Given $\epsilon > 0$, there exists $\delta_1$ such that $|x-a| < \delta_1$ implies that $|f(x) - A| < \epsilon$

Similarly, there exists $\delta_2$ such that $|x-a| < \delta_2$ implies that $|g(x)-B| < \epsilon$

And this is where I got stuck on my proof, I have no idea how to proceed on from here, or with part (ii). Is what I have so far correct? How do I proceed on from here? Any help would be appreciated.

For part 1, consider $\epsilon=\frac{B-A}{2}>0$ and take $\delta=min(\delta_1,\delta_2)$, then you can get your result, since eventually $g(x)-B> \epsilon$ and $f(x) < A+\epsilon$ .
For part 2, consider $f(x)=\frac{(x-a)^2}{2},g(x)=(x-a)^2$ but $A = B=0$