Let $\displaystyle f$ be continuous on $\displaystyle [a,b)$. Prove that$\displaystyle lim_{x \rightarrow b^-}f(x)$ exists $\displaystyle iff$ $\displaystyle f$ is uniformly continuous on $\displaystyle [a,b)$.

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- Aug 11th 2009, 05:14 PMKat-Muniform continuity
Let $\displaystyle f$ be continuous on $\displaystyle [a,b)$. Prove that$\displaystyle lim_{x \rightarrow b^-}f(x)$ exists $\displaystyle iff$ $\displaystyle f$ is uniformly continuous on $\displaystyle [a,b)$.

- Aug 11th 2009, 07:48 PMsiclar
If we know $\displaystyle \displaystyle\lim_{x\rightarrow b-}f(x)$ exists and we want to show f is actually uniformly continuous, it helps to invoke a nice theorem.

**Theorem**: If $\displaystyle f$ is a continuous real function on a closed and bounded set of real numbers, then $\displaystyle f$ is in fact uniformly continuous.

Then if we can define a value of $\displaystyle f$ at $\displaystyle b$ which maintains continuity, we've got it!

__Spoiler__:

Conversely if the function is uniformly continuous then a straightforward use of the definition shows the limit exists.

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