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Find a negative number N

  1. Sep 16, 2009 #1
    Find a negative number N such that [tex]|\frac{x^2}{1+x^2}-1| < \epsilon [/tex] for x < N. That implies [tex] |\frac{-1}{1+x^2}| < \epsilon if x < N[/tex].This gives me the following [tex] -|\frac{1}{1+x^2}| < \epsilon if x < N[/tex].I do not know what to do from here. Please help as I am teaching myself. Thanks.
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  3. Sep 16, 2009 #2


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    Finding "a negative number N such that if x< N then ..." is very peculiar. But since x only occurs to even power, it doesn't matter whether you use x or -x. "Find a positive number M such that if x> N then [itex]|\frac{-1}{1+x^2}|< epsilon[/itex]" is more "standard" and gives N= -M. Saying that [itex]|\frac{-1}{1+x^2}|< \epsilon[/itex] is the same as saying [itex]0< \frac{1}{1+x^2}< \epsilon[/itex] which is, in turn, the same as saying that [itex]1+ x^2> \frac{1}{\epsilon}[/itex] which is the same as [itex]x^2> \frac{1}{\epsilon}- 1[/itex]. Can you carry on from there?
  4. Sep 16, 2009 #3
    [tex] x=+/-\sqrt{\frac{1-\epsilon}{\epsilon}}[/tex] Then I have to determine which one of the roots is the answer, as x is negitive it must [tex]-\sqrt{\frac{1-\epsilon}{\epsilon}}[/tex], but that last answer doesn't sound a very good reason as to why it is the negitive root. I was wondering is there better reasoning?
  5. Sep 16, 2009 #4
    The definition I am using is: Let f(x) be defined for all x in some infinite open interval extending in the negative x-direction. We will write
    [tex]\lim_{x-> -\infty}f(x)=L[/tex]
    if given any number [tex]\epsilon >0 [/tex], there corresponds a negative number N such that
    [tex] |f(x)-L| < \epsilon \mbox{ if } x < N[/tex].
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