Uniform Continuity: Proof of Limit Existence

[Newtime]

Homework Statement

Assume [tex]f:(0,1) \rightarrow \mathbb{R} [/tex] is uniformly continuous. Show that [tex]\lim_{x \to 0^+}f(x)[/tex] exists.

Homework Equations

Basic theorems from analysis.

The Attempt at a Solution

The statement is intuitive but I'm having trouble formalizing the idea. Uniform Continuity means the derivative is bounded. So the function can't veer off to infinity or do something like sin(1/x). But of course, this is flimy reasoning at best. Any ideas are appreciated.

 

[jbunniii]

Claim: Uniform continuity implies the following. For any [itex]\epsilon > 0[/itex], there's an interval [itex]I(\delta) = (0, \delta)[/itex] such that [itex]f(x)[/itex] is within [itex]\epsilon[/itex] of [itex]f(\delta)[/itex] as long as [itex]x \in I(\delta)[/itex].

 

[Newtime]

Claim: Uniform continuity implies the following. For any [itex]\epsilon > 0[/itex], there's an interval [itex]I(\delta) = (0, \delta)[/itex] such that [itex]f(x)[/itex] is within [itex]\epsilon[/itex] of [itex]f(\delta)[/itex] as long as [itex]x \in I(\delta)[/itex].

Thanks for the help. Fortunately, I just solved the question. It's easy once you consider the image of a sequence that converges to 0.