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Epsilon-Delta Proof of limit approaching infinity

  1. Oct 4, 2009 #1
    **DISCLAIMER - I am super bad at LaTeX**

    1. The problem statement, all variables and given/known data

    Prove

    [tex]\lim_{x \rightarrow \infty}\frac{1}{1+x^2} = 0[/tex]


    2. Relevant equations

    I Think I proved it, but I feel like I'm missing something to make this a proof of ALL [tex]\epsilon[/tex]>0 and not just one case. Maybe I did it right. I really don't know. Just looking for a second opinion and/or advice on [tex]\epsilon[/tex]-[tex]\delta[/tex] proofs.

    3.Attempt at a Solution

    the definition logically is if [tex]x > N[/tex] then [tex]|f(x) - L| > \epsilon[/tex] for some [tex]N,\epsilon > 0[/tex]

    Setting [tex]N=\sqrt{\frac{1-\epsilon}{\epsilon}}[/tex]

    x > N

    [tex]\Rightarrow[/tex] [tex] x > \sqrt{\frac{1-\epsilon}{\epsilon}}[/tex]

    [tex]\Rightarrow[/tex] [tex] x^2 > \frac{1-\epsilon}{\epsilon}[/tex]

    [tex]\Rightarrow[/tex] [tex] x^2 + 1 > \frac{1}{\epsilon}[/tex]

    [tex]\Rightarrow[/tex] [tex] \frac{1}{1+x^2} < \epsilon [/tex]

    [tex] \frac{1}{1+x^2} = f(x) [/tex]

    and since N > 0, and x > N, it is implied x > 0 and therefore [tex]|f(x)| = f(x)[/tex]

    I'm not sure if this is a good enough proof? Thanks in Advance :)
     
  2. jcsd
  3. Oct 4, 2009 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, your argument is fine. I would start it and end it slightly differently:

    Begin with:

    Suppose ε > 0 Then your next line

    Let N = ... is OK
    Then, if x > N...
    ...
    ...
    [tex]
    \frac{1}{1+x^2} < \epsilon
    [/tex]

    So

    [tex]
    |\frac{1}{1+x^2}| < \epsilon
    [/tex]

    And stop there.
     
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