Prove $B\ge A$ where $\lim_{x\to x_0}f(x)=A$, $\lim_{x\to x_0}g(x)=B$ and $\exists \delta_1>0$ s.t. $f(x) \le g(x)$

I have the following problem:

If the limits of the functions $f(x)$ and $g(x)$ exist and also there exists a number $\delta_1$ s.t. $\forall x: 0<|x-x_0|<\delta_1$ it is true that $f(x) \le g(x)$.

Then prove that $A \le B$.

What I've got:

Let $\epsilon_1>0, \epsilon_2>0$. Lets denote the limits of $f(x)$ and $g(x)$ A and B. Then there exist numbers $\delta_2, \delta_3$ s.t. $$0<|x-x_0|<\delta_2 \implies |f(x)-A|<\epsilon_1 \implies A-\epsilon_1< f(x) < A + \epsilon_1\\ 0<|x-x_0|<\delta_3 \implies |g(x)-A|<\epsilon_2 \implies B-\epsilon_2< g(x) < B + \epsilon_2.$$

Then when $|x-x_0|<\delta = min(\delta_1, \delta_2, \delta_3)$ it is true that: $$A-\epsilon_1< f(x) \le g(x)<B+\epsilon_2 \implies B + (\epsilon_1 + \epsilon_2)>A.$$

That's where I'm stuck. How do I prove from that, that $B \ge A$?

• You could argue by contradiction: Assume $A>B$, and show that your $\delta_1$ cannot exist, using for instance $\varepsilon = \frac{A-B}2$. Oct 12 '17 at 11:41
• Using that I get that $A-B \le A+B$ which is useless. What conditions should I use for getting a contradiction? Oct 12 '17 at 12:23
• You should get a $\delta_2$ such that $|f(x) - A|<\varepsilon$ and a $\delta_3$ such that $|g(x) - B| <\varepsilon$. This implies that as long as $0<|x-x_0|<\min(\delta_2, \delta_3)$, we must have $f(x)>g(x)$. (Note that what I'm proposing here isn't a way to go on with your proof, but a whole different proof from scratch, which is why I put it in a comment). Oct 12 '17 at 12:27
• @Arthur Got it! Thank you! If you have the time, write an answer, I'll be happy to accept it. Oct 12 '17 at 14:09

What you have so far is fine. Fix $\varepsilon>0$, put $\varepsilon_1=\varepsilon_2=\frac{\varepsilon}{2}$. Then running through your argument, you obtain $A<B+\varepsilon$. Since $\varepsilon>0$ was arbitrary, it follows that $A\leq B$.
• I see it intuitively but how is this written formally? Does $\epsilon \to 0$? Oct 12 '17 at 14:10
• Intuitively yes, $\varepsilon\to0$. Formally, it's impossible to have $A<B+\varepsilon$ for all $\varepsilon>0$ and $B<A$. Oct 12 '17 at 14:15