Limit proof

**Zalren** · November 6th 2010, 03:19 PM

Assume $f(x) >= g(x)$ for all $x$ in some set $A$ on which $f$ and $g$ are defined. Show that for any limit point $c$ of $A$ we must have $\displaystyle \lim_{x \to c} f(x) >= \lim_{x \to c} g(x).$

I think I start with the $\epsilon$ and $\delta$ definition of a limit. So let $\displaystyle \lim_{x \to c} f(x) = F.$ Let $\epsilon > 0.$ $0 < |x - c| < \delta$ implies there exists $\delta > 0$ such that $|f(x) - F| < \epsilon.$

Is that the right place to start? Because I am not sure where to go from here.

Thanks in advance.

**Plato** · November 6th 2010, 03:36 PM

Let $\displaystyle \lim_{x \to c} f(x) =F~\&~ \lim_{x \to c} g(x)=G.$

Suppose that that $F<G$ . You can get a contradiction.

Hint: consider $\varepsilon = \frac{{G - F}}{2}$ .

**Zalren** · November 6th 2010, 04:10 PM

Thank you so much. I believe I have it.

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