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Math Help - Limits of Functions

  1. #1
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    Limits of Functions

    Let f:D-->C(complex numbers).
    Prove that if f(x)-->b as x-->a and c=/b, then there exists a delta>0, such that for any x contained in D, 0<|x-a|<delta ==> |f(x)-c|>(1/2)|b-c|.

    Not really sure how to get started on this one. I'm assuming I need to find a relation between |f(x)-c| and |b-c| but I can't seem to see how to do that.
    Last edited by mr fantastic; October 16th 2010 at 07:02 PM. Reason: Disabled smilies.
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  2. #2
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    You do know the definition of "limit", don't you? If \lim_{x\to a} f(x)= b then, for any \epsilon> 0, there exist \delta> 0 such that if |x- a|< \delta then |f(x)- b|< \epsilon. For any c not equal to b, take \epsilon= |b- c|/2.

    (The fact that this involves complex numbers is irrelevant. Exactly the same proof works if your set is complex numbers or real numbers.)
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  3. #3
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    I did know the definition of limit. The trouble that I was having was with choosing the proper epsilon that makes the proof true. So would you please tell me how you chose the epsilon that you chose?
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  4. #4
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    Let \varepsilon  = \dfrac{{\left| {c - b} \right|}}{2} > 0.
    From the definition  \left( {\exists \delta>0 } \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {b - f(x)} \right| < \varepsilon } \right]

    Suppose that \left( {\exists d} \right)\left[ {\left| {d - a} \right| < \delta \;\& \,\left| {c - f(d)} \right| < \frac{{\left| {c - b} \right|}}<br />
{2}} \right].

    Note that means \left| {c - b} \right| \leqslant \left| {c - f(d)} \right| + \left| {f(d) - b} \right|
    There is a contradiction there. What is it?
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  5. #5
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    well...here is what I see so far...
    Since |b-f(x)|<epsilon and |c-f(d)|<|c-b|/2=epsilon
    then |b-f(x)| + |c-f(d)| should be less then to two times epsilon= |c-b|. But now that I look at this closer I'm not sure that is actually revelant with the statements that you made.
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  6. #6
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    \left| {c - b} \right| \leqslant \left| {c - f(d)} \right| + \left| {f(d) - b} \right|< \frac{|c-b|}{2}+\frac{|c-b|}{2}=|c-b|
    That is a contradiction. No number is less than itself.
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  7. #7
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    Oh... I see it in now. Alright, that makes a lot of sense. Thanks a lot.
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