Uniform Continuity: Proof of Limit Existence

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Homework Statement


Assume f:(0,1) \rightarrow \mathbb{R} is uniformly continuous. Show that \lim_{x \to 0^+}f(x) 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.
 
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Claim: Uniform continuity implies the following. For any \epsilon > 0, there's an interval I(\delta) = (0, \delta) such that f(x) is within \epsilon of f(\delta) as long as x \in I(\delta).
 
jbunniii said:
Claim: Uniform continuity implies the following. For any \epsilon > 0, there's an interval I(\delta) = (0, \delta) such that f(x) is within \epsilon of f(\delta) as long as x \in I(\delta).

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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