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Limits and choosing an epsilon properly?

  1. Oct 25, 2004 #1
    I really need help on solving this question:

    Let d and K be given real numbers. Suppose that lim f(x) > K.
    Show that there is a number h>0 such that f(x) > K for all x in the punctured open interval of width 2h centred at d.

    The only hint that i was given was that if there are two real numbers as close as you like, then they are basically the same real.
    How can i show this using this idea?
  2. jcsd
  3. Oct 25, 2004 #2


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    Dearly Missed

    Is c=d?????????????
  4. Oct 25, 2004 #3
    yes that is the idea, but how am i suppose to show that it's equal
  5. Oct 25, 2004 #4


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    Suppose [tex]\lim_{x\rightarrow c}f(x)=L>K[/tex]

    Write down the definition of the limit in this case. There's a [tex]|f(x)-L|<\epsilon[/tex] part. This controls how close f is to L. By chosing epsilon properly, you can force f to be some distance away from any number not equal to L (on some punctured disc centered at c of course).

    For example if you know [tex]|f(x)-L|<1/2[/tex] on some interval, then can [tex]f(x)=L+1/2[/tex] on this interval? Can it equal anything larger? What's the lowest it could be?
  6. Oct 25, 2004 #5
    What does it mean when asking to chosing an epsilon properly?

    Also, can someone clarify, how the graph of two horizontal lines work? i.e, y=L , y=f.
    Last edited: Oct 25, 2004
  7. Oct 25, 2004 #6


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    You can think of epsilon as bounds for your function on the corresponding interval [tex]0<|x-c|<\delta[/tex].On this interval, your function will only take on values above [tex]L-\epsilon[/tex] and below [tex]L+\epsilon[/tex]. How do you pick epsilon to leave K out of this range? If epsilon is too large, you won't be able to rule out the possibility that f(x)=K.

    Your horizontal lines..y=L would just be a horizontal line at height L, y=f won't necessarily be a horizontal line, f is a function. I don't think I understand your question.
  8. Oct 25, 2004 #7
    So in order to show that f(x)> K for all x.. what are the main steps required in proving this, and also is there any techniques that can be used in solving delta-epsilon type of problems? I m really confused in these types of problems, i dont seem to understand the concept of delta and epsilon, like i kno that they are really small and can be regarded as equal, or not equal but very close, etc.
  9. Oct 26, 2004 #8


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    1) pick an [tex]\epsilon >0[/tex] so that if [tex]|f(x)-L|<\epsilon[/tex] you know [tex]|f(x)-K|>0[/tex]. Your particular epsilon will depend on how far L is from K.
    2) Appeal to the definition of the limit to produce a [tex]\delta >0[/tex] so that if [tex]0<|x-c|<\delta[/tex] then [tex]|f(x)-L|<\epsilon[/tex]. You won't know what this delta is, but the fact that the limit is L guarantees it's existance.
    3) Combine the above.

    epsilon-delta type proofs are important if you want an understanding of calculus. Everything you do later on depends on limits, and without understanding epsilon-delta you won't rigorously understand what a limit is. You really need to work some examples and try to understand the definition of the limit as much as possible. Practice is important.
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