If $f(x)<g(x)$ prove that $\lim f(x)<\lim g(x)$ I have this question:


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


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?
HINT: Suppose that $A&gt;B$, and let $\epsilon=\frac12(A-B)&gt;0$. Show that there is a $\delta&gt;0$ such that $f(x)&gt;A-\epsilon=B+\epsilon&gt;g(x)$ for all $x\in(x_0-\delta,x_0+\delta)$; this contradicts the hypothesis that $f(x)&lt;g(x)$ on an open interval around $x_0$. (Why?)

It’s easy to find examples in which $A=B$. You can do it with $f(x)=0$, in fact.
Quite obvious from the definition of continuity. By continuity,$\forall \epsilon$ there exist $\delta$ s.t $x\in B_{\delta}(x_0) \implies|f(x)-A|&lt;\epsilon$ and $|g(x)-B|&lt;\epsilon$ and so we have $ B-\epsilon&lt;g(x) \ \text{and} f(x)&lt;\epsilon +A$ so $ B&lt;2\epsilon +A \  \forall \epsilon &gt;0$. Since $\epsilon$ is arbitrary so we conclude that $B&lt;A$. But i am not sure what is missing, i am not sure why i couldn't prove that $A=B$