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?

Quite obvious from the definition of continuity. By continuity,$\forall \epsilon$ there exist $\delta$ s.t $x\in B_{\delta}(x_0) \implies|f(x)-A|<\epsilon$ and $|g(x)-B|<\epsilon$ and so we have $ B-\epsilon<g(x) \ \text{and} f(x)<\epsilon +A$ so $ B<2\epsilon +A \  \forall \epsilon >0$. Since $\epsilon$ is arbitrary so we conclude that $B<A$. But i am not sure what is missing, i am not sure why i couldn't prove that $A=B$ 
