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Math Help - Proving limit exists

  1. #1
    Member eXist's Avatar
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    Proving limit exists

    Well the problem reads: Suppose f: (a, b) \rightarrow R and \lim_{x\to\infty}f(x_n) exists for every sequence x_n (where n goes from 1 \to \infty) in (a, b) such that \lim_{x\to\infty}x_n = a. Prove that \lim_{x\to a^+}f(x_n) exists.

    I'm going to assume that \lim_{x\to a^+}f(x_n) does not exist and prove it by contradiction. So then we still have \lim_{x\to\infty}x_n = a. But since the sequence x_n lies in the interval (a, b) the \lim_{x\to\infty}x_n = a can't exist because we assumed that right hand limit: \lim_{x\to a^+}f(x_n) does not exist, so the limit as anything approaches a in the interval (a, b) can't exist. But that is a contradiction since we are given \lim_{x\to\infty}x_n = a.

    I'm not sure if this is right because it seems a little short. Can someone let me know if this is the right approach? Thanks, Chad.

    Edit: Maybe I have to use Bolzano-Weierstrass and say that x_n has a convergent sub sequence that tends to a as x approaches infinity, not really sure that makes a difference though since we already have a sequence that converges to a.
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  2. #2
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    Quote Originally Posted by eXist View Post
    Well the problem reads: Suppose f: (a, b) \rightarrow R and \lim_{x\to\infty}f(x_n) exists for every sequence x_n (where n goes from 1 \to \infty) in (a, b) such that \lim_{x\to\infty}x_n = a. Prove that \lim_{x\to a^+}f(x_n) exists.

    I'm going to assume that \lim_{x\to a^+}f(x_n) does not exist and prove it by contradiction. So then we still have \lim_{x\to\infty}x_n = a. But since the sequence x_n lies in the interval (a, b) the \lim_{x\to\infty}x_n = a can't exist because we assumed that right hand limit: \lim_{x\to a^+}f(x_n) does not exist, so the limit as anything approaches a in the interval (a, b) can't exist. But that is a contradiction since we are given \lim_{x\to\infty}x_n = a.

    I'm not sure if this is right because it seems a little short. Can someone let me know if this is the right approach? Thanks, Chad.

    Edit: Maybe I have to use Bolzano-Weierstrass and say that x_n has a convergent sub sequence that tends to a as x approaches infinity, not really sure that makes a difference though since we already have a sequence that converges to a.
    You have to assume that the limit c=f(x_n) is the same for any sequence x_n \rightarrow a (otherwise the conclusion would not be valid for obvious reasons). Assume that \lim_{x\rightarrow a^+ } f(x) doesn't exists, in particular there would exist an \epsilon >0 such that for every 0<\delta (=1/n) we have x_{\delta }-a < \delta and \vert f(x_{\delta })-c \vert \geq \epsilon

    Edit: Your argument fails because you assume that because \lim_{x\rightarrow a^+ } f(x) doesn't exist this implies that the limit f(x_n) doesn't exist for any sequence converging to a.
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  3. #3
    Member eXist's Avatar
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    So if I assume that the limit: \lim_{x\to a^+}f(x) does not exist, that changes the fact that every sequence converges to a? I might be miss understanding you.

    I understand that a must be a cluster point since every sequence in the interval (a, b) converges to it, and therefor the limit at a from the right must exist since the terms of the sequence are all larger than a since they are in the interval (a, b).

    I see why this is true, I'm just struggling to write it down I guess.
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