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Proving limit statement using delta-epsilon definition

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    For all [itex]x\in R[/itex] , [itex]f(x)>0[/itex] . Using precise definition of limits and infinite limits, prove that [itex]\lim_{x\to a}f(x)=\infty[/itex] if and only if [itex]\lim_{x\to a}\frac{1}{f(x)}=0 [/itex]

    2. Relevant equations



    3. The attempt at a solution

    I know the precise definition of limits and infinite limits but I cannot see how they can be applied in this case. Also this is biconditional statement so I guess I gotta deduce 2 sub-proofs before drawing the conclusion. Ok that's all I've thought of this far, any hints would be greatly appreciated, thanks!
     
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  3. Sep 10, 2012 #2

    SammyS

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    If [itex]\displaystyle \lim_{x\to\,a}\frac{1}{f(x)}=0 \,,[/itex] then restate this using δ - ε language.

    That should get you started.
     
  4. Sep 10, 2012 #3
    Ok here's my working for the first part, that is if [itex]\lim_{x\to a}\frac{1}{f(x)}=0 [/itex] then [itex]\lim_{x\to a}f(x)=\infty[/itex].

    Since [itex]\lim_{x\to a}\frac{1}{f(x)}=0 [/itex] exists,
    if we let ε > 0, then there exists a δ such that
    [itex]0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} < \varepsilon[/itex]

    Choose [itex]M=\frac{1}{ε}[/itex] > 0, then
    [itex]0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} < \frac{1}{{{M}}}\\
    0 < \left| {x - a} \right| < \delta \Rightarrow \left|f(x)\right| > M[/itex]

    How could I get rid of the modulus of [itex]f(x)[/itex] in this case?
     
    Last edited: Sep 11, 2012
  5. Sep 11, 2012 #4

    SammyS

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    It's almost there.

    (I'm still talking about the first part of the proof here.)

    What is the definition for [itex]\lim_{x\to a}f(x)=\infty\ ?[/itex] In other words how do we show that [itex]\lim_{x\to a}f(x)=\infty\ ?[/itex]

    You need to show that, given any M > 0, there exists a δ such the f(x) > M, whenever 0 < |x-a| < δ .

    So, to begin the proof, let M > 0.

    Then let ε = 1/M.

    Since you are assuming that [itex]\displaystyle \lim_{x\to a}\frac{1}{f(x)}=0 [/itex], you can get your δ which makes [itex]\displaystyle \frac{1}{f(x)}<\varepsilon= \frac{1}{M} \ \ \dots[/itex]
     
  6. Sep 11, 2012 #5
    When it comes to this kind of problem I always try to manipulate f(x) to the form of |x-a| before choosing a δ that works. But in this generalized case how am I supposed to do that?
     
  7. Sep 11, 2012 #6
    Drawar, the technique of manipulating the form of |f(x) - L| and so on is a good one for trying to estimate the [itex]\epsilon[/itex]'s and [itex]\delta[/itex]'s you need. But eventually you have to understand what your goal is, and what you need to find.

    As SammyS said:
    You begin by having a general M > 0, for which you need to find a suitable [itex]\delta[/itex], so that for each x satisfying [itex]|x-a| < \delta[/itex], f(x) > M holds.
    But f(x) > M means: [itex]\frac{1}{f(x)}< \frac{1}{M}[/itex]. Can you find such a [itex]\delta[/itex] that will ensure that? I bet you can -- it's very much implied from your assumption on [itex]\frac{1}{f(x)}[/itex]....
    Hope this helps!
     
  8. Sep 12, 2012 #7
    But how should we choose such a δ? In terms of ε or M? Sorry you guys but I don't really get what you said. I thought after letting ε=1/M then the proof is completed.
     
  9. Sep 12, 2012 #8

    SammyS

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    Well, this is a different sort of problem.

    You're not give a specific function to manipulate in the way you describe.

    You assume that [itex]\displaystyle \lim_{x\to\,a}\frac{1}{f(x)}=0[/itex] is true.

    Now to show that [itex]\displaystyle \lim_{x\to\,a}f(x)=\infty\ ,[/itex] you need to prove that given any M > 0, (no matter how large) , there exists a δ such that f(x) > M whenever 0<|x-a|<0 .

    Use 1/M as your ε. Then using this 1/M as ε with [itex]\displaystyle \lim_{x\to\,a}\frac{1}{f(x)}=0[/itex] find a delta. Show the this δ works with M for [itex]\displaystyle \lim_{x\to\,a}f(x)=\infty\ .[/itex]
     
    Last edited: Sep 12, 2012
  10. Sep 12, 2012 #9
    Well - that's exactly it! You need to promise me you can find such a delta - even without specifying it!
    So by choosing, as you and others mentioned, ε = 1/M, you can say, according to your assumption, that there exists a δ (let's call it δ'!), so that for every x satisfying |x-a| < δ',
    [itex]\frac{1}{f(x)} < \epsilon[/itex] holds...
    This δ' is what you need....
     
  11. Sep 12, 2012 #10
    So the (<=) proof would look like this, right?

    Let M > 0 and let ε=1/M > 0
    Since [itex]\lim_{x\to a}\frac{1}{f(x)}=0 [/itex] exists, then there exists a δ such that
    [itex]0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{\left| {f(x)} \right|}} < \varepsilon[/itex]
    [itex]0 < \left| {x - a} \right| < \delta \Rightarrow \frac{1}{{ {f(x)}}} < \frac{1}{{{M}}}\\
    0 < \left| {x - a} \right| < \delta \Rightarrow f(x) > M[/itex]
    this means [itex]\lim_{x\to a}f(x)=\infty[/itex]
     
    Last edited: Sep 12, 2012
  12. Sep 12, 2012 #11
    Looks perfect to me ;)
    You could sum up by saying: for every M > 0 we have therefore found a δ > 0, so that for every x satisfying |x - a| < δ, f(x) > M holds - which makes it perfect and shows that you know what you're doing - but the important thing is that you understood the logic of it!
     
  13. Sep 12, 2012 #12
    Hell yeah thank you so much!!!
    Ok for the (=>) proof, I think just let M=1/ε and do everything backwards. Am I right?
     
  14. Sep 12, 2012 #13
    Yup, sounds about right ;) Make sure you formulate it logically and you have it...
     
  15. Sep 12, 2012 #14
    Thank you guys for all of your kind help and guidance. :) Finally figured it out now.
     
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