# Limit Analysis Question

• Nov 24th 2008, 05:52 AM
Uriah
Limit Analysis Question
Hey.

I have played around with this question for my analysis class and I would like some help:

Given: L= lim f1(x), M=lim f2(x)

Show that if f1<=f2 for all x in some interval (a,b), then L<=M
• Nov 24th 2008, 06:07 AM
Opalg
It's probably easiest to do this by contradiction. Suppose that L>M and let $\varepsilon = \tfrac12(L-M)$. For x sufficiently close to some limit point c (the limit point isn't specified in the question), it will be true that $|f_1(x)-L|< \varepsilon$ and $|f_2(x)-M|< \varepsilon$. From that, and the fact that $f_1(x)\leqslant f_2(x)$, you should be able to use the triangle inequality to get a contradiction.
• Nov 24th 2008, 07:57 AM
Uriah
I am still struggling here. I can't find a way to make the triangle inequality work for this because if I use it on $|f_1(x)-L|< \varepsilon$ or $|f_2(x)-M|< \varepsilon$ I know nothing about the relationship between $|f_1(x)| + |-L|$ and epsilon.
• Nov 24th 2008, 08:53 AM
Opalg
Quote:

Originally Posted by Uriah
I am still struggling here. I can't find a way to make the triangle inequality work for this because if I use it on $|f_1(x)-L|< \varepsilon$ or $|f_2(x)-M|< \varepsilon$ I know nothing about the relationship between $|f_1(x)| + |-L|$ and epsilon.

Okay, $|f_1(x)-L|< \varepsilon$ is equivalent to $L-\varepsilon. Similarly, $M-\varepsilon. From those, you can extract $L-\varepsilon. Can you finish it from there?
• Nov 24th 2008, 09:42 AM
Uriah
Duh. I can't believe I didn't look at it like that. Thanks!