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please I require help in showing that if $f$ is uniformly continuous on a bounded interval $(a,b)$, then $\lim_{x\to b^-} f(x)$ exists.

edit:

$f$ is uniformly continuous on $(a,b)$ implies that for every $\epsilon \gt 0$ there is a $\delta \gt 0 $ such that $|f(x) -f(y)| \lt \epsilon$ whenever $|x-y|\lt \delta$ for every $x,y$.

let $x_n$ and $y_n$ be sequences in $(a,b)$ that converge to $b$. Then there is a natural number $N$ such that for every $n \geq N$, $0<b-x_n< \delta/2$ and $0<b-y_n< \delta/2$ so that $$ |x_n-y_n| \leq |x_n -b| + |y_n -b| \lt \delta.$$ So, $|f(x_n) -f(y_n)| \lt \epsilon$ and this implies that $\lim f(x_n) = \lim f(y_n)$.

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2 Answers

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Expanding azarel's hint:

1) Show that a uniformly continuous function maps Cauchy sequences to Cauchy sequences. This isn't hard to do (and is a very useful property), it follows almost directly from the relevant definitions.

2) Take a fixed sequence $(x_n)$ in $(a,b)$ converging to $b$. By the first step, $\bigl(f(x_n)\bigr)$ is Cauchy and thus converges to some number $\alpha$.

3) Now suppose that $\lim\limits_{x\rightarrow b^-} f(x)\ne\alpha$. Then you can find an $\epsilon>0$ and a sequence $(y_n)$ in $(a,b)$ converging to $b$ such that for all $n$, $|f(y_n)-\alpha|>\epsilon$.

4) Consider what we have: $(x_n)$ and $(y_n)$ are both sequences in $(a,b)$ converging to $b$. On one hand we have that $|x_n-y_n|$ can be made as small as you like. But on the other hand, for large $n$, the quantity $|f(x_n)-f(y_n)|\approx|\alpha-f(y_n)|$ is big. Could this occur for a function that is uniformly continuous on $(a,b)$?

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thanks for your post. why is $|f(x_n)-f(y_n)|\approx|\alpha-f(y_n)|$? –  joel Mar 13 '12 at 18:51
    
@joel by 2), $f(x_n)\approx \alpha$ (the sequence $(f(x_n))$ converges to $\alpha$). –  David Mitra Mar 13 '12 at 18:53
    
Thanks again: So going through steps 1 through to 4 will establish the proof? –  joel Mar 13 '12 at 18:58
    
@joel Yes. Or, you can just prove 1). Then, the argument in your post together with 1) will imply that there is a $L$ such that for each sequence $(x_n)$ in $(a,b)$ converging to $b$, the sequence $(f(x_n))$ converges $L$. This implies that $\lim_{x\rightarrow b^-} f(x)$ exists. –  David Mitra Mar 13 '12 at 19:05
    
Thanks very much. –  joel Mar 13 '12 at 19:12
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$\bf Hint:$ Show that if $(x_n)\subseteq (a,b)$ converges to $b$ then $(f(x_n))$ is a Cauchy sequence and then show that the limit does not depend on the chosen sequence.

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I have added to my post. What else needs to be done? Thanks. –  joel Mar 13 '12 at 17:42
    
You are only left to show that $(f(x_n))$ is a Cauchy sequence. –  azarel Mar 13 '12 at 17:54
    
Okay..I can show that $((fx)n))$ is a Cauchy sequence. but how does that show that $\lim_{x\to b^-}$ exists? –  joel Mar 13 '12 at 18:17
    
@joel Since $\mathbb R$ is complet every Cauchy sequence converges. –  azarel Mar 13 '12 at 21:29
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