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I can see that $f(x_n)$ converges, since it is bounded and monotonic. Let me give some context and mention that we have dwelled mostly on the sequential definition of limits. Does $(x_n)$ need to be nondecreasing as well? I suppose I need to use sups and infs somehow, but could use a hint. This is not homework but a rather an optional practice problem to prepare for a test.

NS248
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  • This is a one-limit from the left. I forgot to mention this above. – NS248 Mar 18 '14 at 03:53
  • My problem is that I need the monotonicity of $(x_n)$ for the limit of $f(x_n)$ to exist in the first place. Therefor I don't see how to use tHe monotone subsequence of $(x_n)$ to my advantage, since the definition requires every sequence less than $b$ to get the job done. – NS248 Mar 18 '14 at 20:48

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Since $f$ is bounded we have $|f(x)|<M$ for some $M$ on $(a,b)$ Consider the set $\{f(x) : x<b\}$, this set is bounded above by $M$ thus it has a supremum $L$, that is $L=\sup\{f(x) : x<b\}$. I claim that $\displaystyle\lim_{x\to b^-}f(x)=L$. Let $\varepsilon>0$, by the definition of the $\sup$ there is a $y\in (a,b)$ such that $L-\varepsilon < f(y) \le L$ .Take $x \in (b-\delta,b)$ then $L-\varepsilon <f(y)\le f(x) \le L$ which means that $|f(x)-L|<\varepsilon$ when $0<b-x<\delta$, therefore $\displaystyle\lim_{x\to b^-}f(x)=L$.

Try finding the $\delta$.

user10444
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Hint: Choose any monotonically increasing sequence $x_n \to b$; then since $f$ is nondecreasing, $f(x_n)$ is a monotonically increasing and bounded sequence, with limit $L$.

Now take any sequence $y_n$ converging to $b$; can you convince yourself that $f(y_n) \to L$ as $y_n \to b$? Choosing a monotonically increasing subsequence $y_{n_k}$ may be of use to you.

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    Every sequence has a monotonic subsequence and subsequences have the same limit as the mother sequence. Now I can use the monotonicity of the subsequence. In other words, thanks for excellent hint. – NS248 Mar 18 '14 at 03:59
  • @senselessapprentice You're very welcome. –  Mar 18 '14 at 04:18
  • Could one construct this monotonic subsequence ? Let $(y_n) \to b$ be the sequence with $y_n \lt b$. We can set $x_1 = y_1$ and get $x_{n+1}$ from the $(b - x_n)/2$ neighborhood of $b$. Does this sound like a good approach? – NS248 Mar 18 '14 at 04:32
  • That notation is less than ideal. I suppose I should have used nested subscripts and just $y$, but perhaps you see what I am getting at. – NS248 Mar 18 '14 at 04:34
  • I unchecked this question as answered not out of ingratitude for your help but only to get additional help as new doubts trouble me. I can't see how the monotonic subsequence can help me, though I believe that it can. But I can't see the jump. The limit has to hold for every sequence. Do I have to prove that every such sequence is eventually monotonic? – NS248 Mar 18 '14 at 20:40