If $f(x)I have this question:

Let $f(x)→A$ and $g(x)→B$ as $x→x_0$. Prove that if $f(x) < g(x)$ for all $x∈(x_0−η, x_0+η)$ (for some $η > 0$) then $A\leq B$. In this case is it always true that $A < B$?

I've tried playing around with the definition for limits but I'm not getting anywhere. Can someone give me a hint on where to start?

This is a standard problem using $\epsilon-\delta$ definition. You can assume $B=0$ by replacing $f'=f-g,g'=g-g=0$. Then you are in the situation that $f'<0$ for all $x$ in $x_{0}$'s small enough neighhorhood, then $\lim_{x\rightarrow x_{0}}f'<0$ as well. It is not difficult to construct a proof based on contradiction. For example consider a translation of above statement into $f'<\epsilon$ in the neighorhood for any given $\epsilon>0$. Then if $\lim_{x\rightarrow x_{0}}f'>0$, what does this tell you?
