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Homework Help: Limit Proof

  1. Mar 24, 2008 #1
    1. The problem statement, all variables and given/known data[/bProve
    Prove: If the limit as x goes to a of f(x)=infinity, then lim as x goes to a of 1/(f(x) =0


    2. Relevant equations
    Need to show with a delta-epsilon proof

    3. The attempt at a solution

    using the definition, lim as x goes to a f(x)=infinity means that for any M>0 there exists an delta>0, where a<x<a+delta implies that f(x)>M. So using this def, I know there is M>0, there exists a delta (not sure what yet) so that a<x<a+delta and taking 1/f(x) shifts the bounds a+delta<x<a and then M would be less than or equal to 0 therefore the lim as x goes to a of 1/f(x)=0.

    Am I close? Please help a girl in distress! lol
    Michelle
     
  2. jcsd
  3. Mar 24, 2008 #2
    well, what you need to show that, [tex]\lim_{x\rightarrow a}\frac{1}{f(x)}=0[/tex], in epsilon delta we have: [tex]\forall\epsilon>0,\exists\delta>0, such \ \ that \ \ |\frac{1}{f(x)}|<\epsilon [/tex].

    using the fact that [tex] \lim_{x\rightarrow a}f(x)=\infty[/tex], as u stated, with a little omission, means that: For any M>0, [tex]\exists\delta>0[/tex] such that whenever

    [tex]0<|x-a|<\delta[/tex] we get f(x)>M. NOw using this fact here we get that whenever

    [tex]0<|x-a|<\delta[/tex] we have [tex]\frac{1}{f(x)}<\frac{1}{M}=\epsilon=>|\frac{1}{f(x)}|<\epsilon=\frac{1}{M}[/tex]

    Now you only need to put everything together, because it is a little messy.

    I hope this helps. But feel free to ask again, don't put too much stress on yourself!
     
    Last edited: Mar 24, 2008
  4. Mar 24, 2008 #3
    i understand but what is [tex]? at the top, I am not sure what that is.
    thanks.
     
  5. Mar 24, 2008 #4
    I don't know what are u talking about? I just used latex to write those math symbols, probbably your browser, or sth, is not being able to generate those symbols, because i do not see [tex] anywhere neither at the top nor at the bottom.
     
    Last edited: Mar 25, 2008
  6. Mar 25, 2008 #5
    that thing about shifting the bounds doesn't really make sens to me.
     
  7. Mar 25, 2008 #6
    it is something our teacher showed us how to do, but it never seems to work for me. One more question, we (meaning you) showed the limit is equal to zero becasue 1/M <epsilon? I get confused because we let m>0. so how can the limit be =0.
     
    Last edited: Mar 25, 2008
  8. Mar 25, 2008 #7
    We, like i wrote before on my other post: We need to show that [tex]\lim_{x\rightarrow a}\frac{1}{f(x)}=0[/tex], right?? using the fact that is in there. So using epsilon delta language, it actually means that we need to show that: [tex]\forall\epsilon>0,\exists\delta>0, such \ \ that \ \ |\frac{1}{f(x)}|<\epsilon [/tex]
    Right?

    Now from:
    [tex] \lim_{x\rightarrow a}f(x)=\infty[/tex]
    we know that for every M>0 ,[tex]\exists\delta>0[/tex] such that for every [tex]xE(a-\delta,a+\delta),[/tex] with the possible exception when x=a, we have [tex]f(x)>M[/tex], but also for that delta and for that interval we also have [tex]\frac{1}{f(x)}<\frac{1}{M}=\epsilon=>|\frac{1}{f(x )}|<\epsilon=\frac{1}{M}[/tex]

    so our line at the top is fullfiled now, because we found that for every epsilon, and also for [tex]\epsilon=\frac{1}{M},\exists\delta(\epsilon)>0,[/tex] such that whenever, [tex]0<|x-a|<\delta=>|\frac{1}{f(x)}|<\frac{1}{M}=\epsilon [/tex] this is all we needed to show.
     
  9. Mar 25, 2008 #8
    got it thanks
     
  10. Mar 25, 2008 #9
    Good job!



    No, problem.
    I love the idea behind this forum. It is blessed.
     
  11. Mar 25, 2008 #10
    yes it is!
     
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