Uniform Continuity: Proof of Limit Existence

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SUMMARY

The discussion centers on proving the existence of the limit \(\lim_{x \to 0^+} f(x)\) for a uniformly continuous function \(f: (0,1) \rightarrow \mathbb{R}\). It is established that uniform continuity ensures the function does not diverge, as it implies bounded derivatives. The key insight is that for any \(\epsilon > 0\), there exists an interval \(I(\delta) = (0, \delta)\) where \(f(x)\) remains within \(\epsilon\) of \(f(\delta)\), leading to the conclusion that the limit exists as \(x\) approaches 0 from the right. The problem is resolved by considering the behavior of sequences converging to 0.

PREREQUISITES
  • Understanding of uniform continuity in real analysis
  • Familiarity with limits and convergence of sequences
  • Basic knowledge of derivatives and their properties
  • Proficiency in mathematical proofs and theorems
NEXT STEPS
  • Study the implications of uniform continuity on function behavior
  • Explore the concept of limits in real analysis
  • Learn about the properties of bounded derivatives
  • Review sequences and their convergence criteria
USEFUL FOR

Students of real analysis, mathematicians focusing on continuity and limits, and anyone interested in formal proofs related to function behavior in calculus.

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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.
 

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