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I have to proof that $\lim_{x \to \c} frac{1}{f(x) = 0$

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Given is: f is a function that maps D onto the Real numbers, and c is within D and it is a limit point, and f(x) =/= 0 for all x in D, and [itex]\lim_{x \to c} f(x) = \infty[/itex]
    I have to proof that:
    [itex]\lim_{x \to c} \frac{1}{f(x)} = 0[/itex]

    3. The attempt at a solution

    This means that according to the definition I have to proof that [itex]\forall \epsilon \ \exists \delta[/itex] so that [itex] \forall x \in D \ with \ 0 < |x-c|< \delta \ \ : |\frac{1}{f(x)} - 0|< \epsilon [/itex].

    Im not sure how to go on from here. Or do i have to do something else?
     
  2. jcsd
  3. Dec 8, 2012 #2

    jbunniii

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    Hint:
    [tex]\left|\frac{1}{f(x)} - 0\right| < \epsilon[/tex]
    is equivalent to
    [tex]|f(x)| > \frac{1}{\epsilon}[/tex]
     
  4. Dec 8, 2012 #3
    Would it then be enough to say that because f diverges to infinity as x approaches c, this by definition means that there must be a [itex]\delta[/itex] for which if [itex]0<|x-c|<\delta[/itex] the latter inequality that you wrote holds, which is equivalent to the former, quod erat demonstrandum?
     
    Last edited: Dec 8, 2012
  5. Dec 8, 2012 #4

    SammyS

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    What you have to prove is what you have in the following statement .
    What will help you get there is to state what is meant by:

    [itex]\displaystyle \lim_{x \to c} f(x) = \infty\ .[/itex]
     
  6. Dec 9, 2012 #5
    Well the weird thing is that in our book we hv only defined what is meant by a sequence diverging to infinity at a point c. This assignment is part of the chapter on continuity where weve only defined what is meant by a limit of f as x goes to infinity. But wut i wrote should b ok since that follows from the definition right?
     
  7. Dec 9, 2012 #6

    SammyS

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    What you wrote is what needs to be proved.

    What you need, among other things, to get there is a definition of what it means for [itex]\displaystyle \lim_{x \to c} f(x) = \infty\ .[/itex]

    You may be able to deduce this from
    what is meant by a sequence diverging to infinity at a point c.​
    and from
    what it means for the limit of a function, f(x), to converge as x → c . ​

    Basically, what it means for [itex]\displaystyle \lim_{x \to c} f(x) = \infty\,,[/itex] is that given any M > 0, (usually M is a large number) there exists a δ > 0 such that for any x satisfying 0<|x-c|<δ, you have that f(x) > M .
     
  8. Dec 9, 2012 #7
    No but i ment what i wrote in reply to the firs post. There must be a delta greater then 1/epsilon
     
  9. Dec 9, 2012 #8

    SammyS

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    That works, if you spell out why it works.

    Given an ε > 0, then 1/ε > 0 . Since [itex]\displaystyle \ \lim_{x \to c} f(x) = \infty\ \ [/itex] there exists δ > 0, such that ... f(x)>1/ε ...

    Added in Edit:

    I removed the absolute value from f(x) above
     
    Last edited: Dec 9, 2012
  10. Dec 9, 2012 #9
    OK thank you. Yeah I know that in writing a proof one has to write a lot of little things to make it formally correct, I just wanted to know if I understand the method of proofing. Thanks again : )
     
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