# Limit Analysis Question

• Nov 24th 2008, 04:52 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, 05:07 AM
Opalg
It's probably easiest to do this by contradiction. Suppose that L>M and let $\displaystyle \varepsilon = \tfrac12(L-M)$. For x sufficiently close to some limit point c (the limit point isn't specified in the question), it will be true that $\displaystyle |f_1(x)-L|< \varepsilon$ and $\displaystyle |f_2(x)-M|< \varepsilon$. From that, and the fact that $\displaystyle f_1(x)\leqslant f_2(x)$, you should be able to use the triangle inequality to get a contradiction.
• Nov 24th 2008, 06:57 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 $\displaystyle |f_1(x)-L|< \varepsilon$ or $\displaystyle |f_2(x)-M|< \varepsilon$ I know nothing about the relationship between $\displaystyle |f_1(x)| + |-L|$ and epsilon.
• Nov 24th 2008, 07:53 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 $\displaystyle |f_1(x)-L|< \varepsilon$ or $\displaystyle |f_2(x)-M|< \varepsilon$ I know nothing about the relationship between $\displaystyle |f_1(x)| + |-L|$ and epsilon.

Okay, $\displaystyle |f_1(x)-L|< \varepsilon$ is equivalent to $\displaystyle L-\varepsilon<f_1(x)<L+\varepsilon$. Similarly, $\displaystyle M-\varepsilon<f_2(x)<M+ \varepsilon$. From those, you can extract $\displaystyle L-\varepsilon<f_1(x)\leqslant f_2(x)<M+ \varepsilon$. Can you finish it from there?
• Nov 24th 2008, 08:42 AM
Uriah
Duh. I can't believe I didn't look at it like that. Thanks!