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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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    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.2011-11-04

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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 $g(x) > f(x)$.