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Homework Help: Yet another epsilon-delta proof

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that [tex]\lim_{x \to -\infty} \frac{2}{\sqrt{x^{4}+1}}=0\[/tex]

    2. Relevant equations

    3. The attempt at a solution
    Preliminary Work
    http://img339.imageshack.us/img339/401/proofzr7.jpg [Broken]

    Before proceeding on with the proof, when we look at the last line there seems to be a logical problem. We know that [tex]\epsilon[/tex] > 0 and x < N < 0 . Thus, it seems counter-intuitive that x is greater than a positive expression but less than a negative. Just a guess but I think it has something to deal with the second last line involving the square root. Thanks in advance!

    Edit: Are there any texts that you guys suggest for learning epsilon-delta proofs? The examples that we do in class seem to be repetitive but when it becomes more abstract and general, some ingenuinity is needed and I would like to see some of these proofs worked out. Thanks!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 17, 2007 #2
    Should move this to calculus for better results and don't expect much. Delta-epsilon proof is a confusing topic
  4. Oct 17, 2007 #3


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    It sure is counterintuitive. How can you say that the solution set to |x|>C is C<x<-C? Don't you mean x<-C or x>C?? E.g. imagine C=2.
  5. Oct 17, 2007 #4
    How do i get from the penultimate line to a solution for x? If I proceed:
    x < [tex]\sqrt[4]{\frac{2}{\epsilon^{2}}-1}[/tex] and set N to that when we give the formal proof, this would contradict one of our initial statements that N < 0 since the expression [tex]\sqrt[4]{\frac{2}{\epsilon^{2}}-1}[/tex] must be positive. So I'm not sure what the solution set of x would be.
    Last edited: Oct 17, 2007
  6. Oct 17, 2007 #5
    Can you use what you know about x<0? So from your next to last line, say
    [tex]|x| >\sqrt[4]{\frac{2}{\epsilon}-1}[/tex]
    [tex]-|x|<\sqrt[4]{\frac{2}{\epsilon}-1}< |x|[/tex], but you know that x=-|x|

    Okay, I obviously need to work on my tex communication skills, but hopefully you can make sense of this.
    Last edited: Oct 17, 2007
  7. Oct 17, 2007 #6


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    Call C your fourth root expression. Yes, C is positive. You want |x|>C and x negative. You want x<-C. I don't see your 'contradictions'.
  8. Oct 17, 2007 #7
    Ah it's all clicked in guys. So for the formal proof I'll set N = -C.

    Edit: Sorry if I need to post another thread but just a quick question about a simpler proof.
    Prove that [tex]\lim_{x \to -\infty} \frac{1}{x}=0[/tex]

    In the solutions manual:
    For x < 0, [tex]|\frac {1}{x}-0|=- \frac{1}{x}[/tex]. If [tex]\epsilon[/tex] > 0 is given, then [tex]- \frac {1}{x}< \epsilon[/tex] which means x < [tex]- \frac{1}{\epsilon}[/tex].

    Take N = [tex]- \frac{1}{\epsilon}[/tex]. Then,

    [tex]x < N[/tex]
    [tex]x < -\frac{1}{\epsilon}[/tex]
    [tex]|-\frac{1}{x} - 0| = -\frac{1}{x} < \epsilon[/tex]
    So, [tex]\lim_{x \to -\infty} \frac{1}{x}=0[/tex]

    My problem is with the initial statement claiming that [tex]|\frac {1}{x}-0|=- \frac{1}{x}[/tex]. I fail to see how this works as the absolute value of a function yields positive values, does it not? And wouldn't the proof work if we just made the condition that [tex]\frac{1}{x} < 0 < \epsilon[/tex] , implying that [tex]x < 0 < \frac{1}{\epsilon}[/tex]?

    I apologize in advance if my train of thought isn't presented very clearly. Thanks again!
  9. Oct 17, 2007 #8
    Again, your limit is as x approaches neg inf, so x is negative. So the absolute value |1/x -0| = -1/x is positive.
  10. Oct 17, 2007 #9
    Also, since x<0, when you multiply both sides of 1/x < eps by x, the inequality flips. Does that make sense? You know that eps is positive and x is negative, so you can't say that 1/eps < x. Sorry for my lack of tex. I'm working on that.
  11. Oct 17, 2007 #10
    Ah It all fell into place. Thanks a lot klile82 and Dick!

    I'll definitely be back with more ... XD
  12. Oct 18, 2007 #11
    Following Dick's advice and posting a new thread.
    Last edited: Oct 18, 2007
  13. Oct 18, 2007 #12


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    The statement you want to prove doesn't make much sense. Could you post it exactly as stated? And I would suggest you start another thread with a new problem. You'll get a lot more attention that way. If I see an active thread with over 5 posts, I don't generally don't even check it. Figure the issue is in the process of being beaten to death. Or that the poster is so confused no one can help.
  14. Oct 18, 2007 #13
    Sure no problem. And i wrote the question wrong, sorry!
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