Prove: 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 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

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