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I've been self studying real analysis using Gaughan's Introduction to Analysis. In the chapter on the algebra of limits of functions he gives the following theorem, leaving the proof as an exercise:

Suppose $f:D \to\mathbb R$ and $g:D \to \mathbb R$, $x_0$ is an accumulation point of $D$ and $f$ and $g$ have limits at $x_0$. If $f(x) \leq g(x)$ for all $x \in D$, then $\lim \limits_{x\to+x_0}f(x) \leq \lim \limits_{x \to +x_0} g(x)$.

I've been thinking about this for a couple of days and can't seem to make any progress. I tried using the formal $\delta$-$\epsilon$ definition and thought about modeling the proof after the proof of the squeeze theorem since the two are kind of similar but nothing has come of it. Any hints as to how to approach this would be appreciated!

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By subtracting, enough to show if $0 \leq h(x)$ for all $x$ then $0 < \lim_{x \rightarrow x_0} h(x) = L$. But if $L < 0$ then can't you get to within $|L/2|$ of it by terms of $h$ (and hence be less than zero)? –  tkr Nov 4 '11 at 2:51
    
CritChamp a good approach is to prove initially that if $g(x)\geq 0$ and there exists $\lim_{x\to x_{0}}g(x)$ then $0\leq $\lim_{x\to x_{0}}g(x)$. This result of course implies what you want and makes $\varepsilon \delta$ argument more manageable. –  Leandro Nov 4 '11 at 2:55
    
Thanks to Bill Cook for fixing some formatting issues. Thank you for your suggestions. As soon as my headache goes away I'll get back to it... –  CritChamp Nov 4 '11 at 3:43
    
This must be a great place - thinking of your history here, you seems to be very grateful. –  AD. Nov 4 '11 at 5:29
    
CritChamp: you have asked 6 questions (including this) and accepted answers for none. Were the answers to your questions satisfactory? If you did find the answers helpful, please consider accepting them. –  Srivatsan Nov 4 '11 at 15:10

2 Answers 2

Hint

If you look at $h=g-f$, then $h\ge0$ on $D$.

  1. Prove that $$\lim_{x\to x_0}h(x) \qquad\text{ exists.}$$

  2. Prove that $$\lim_{x\to x_0}h(x)\ge 0.$$

  3. Use that that $$\lim_{x\to x_0}h(x) = \lim_{x\to x_0}g(x) - \lim_{x\to x_0}f(x).$$

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It's probably easier to prove the contrapositive. In other words, if $\lim_{\rightarrow x_0+} g(x) > \lim_{\rightarrow x_0+} f(x)$, then show there is at least one $x$ for which $g(x) > f(x)$.

Define $A = \lim_{\rightarrow x_0+} g(x)$ and $B = \lim_{\rightarrow x_0+} f(x)$. If $A > B$, you can just choose some $C$ with $A > C > B$. Then for $x$ close enough to $x_0$ you have both $f(x) < C$ and $g(x) > C$. This means $g(x) > f(x)$.

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