1
$\begingroup$

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 $B \ge A$?

$\endgroup$
  • 1
    $\begingroup$ You could argue by contradiction: Assume $A>B$, and show that your $\delta_1$ cannot exist, using for instance $\varepsilon = \frac{A-B}2$. $\endgroup$ – Arthur Oct 12 '17 at 11:41
  • $\begingroup$ Using that I get that $A-B \le A+B$ which is useless. What conditions should I use for getting a contradiction? $\endgroup$ – Nikola Oct 12 '17 at 12:23
  • 1
    $\begingroup$ 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). $\endgroup$ – Arthur Oct 12 '17 at 12:27
  • $\begingroup$ @Arthur Got it! Thank you! If you have the time, write an answer, I'll be happy to accept it. $\endgroup$ – Nikola Oct 12 '17 at 14:09
1
$\begingroup$

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 $A\leq B$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see it intuitively but how is this written formally? Does $\epsilon \to 0$? $\endgroup$ – Nikola Oct 12 '17 at 14:10
  • $\begingroup$ Intuitively yes, $\varepsilon\to0$. Formally, it's impossible to have $A<B+\varepsilon$ for all $\varepsilon>0$ and $B<A$. $\endgroup$ – Aweygan Oct 12 '17 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.