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Limits, find the smallest +ve number N

  1. Sep 15, 2009 #1
    Find the smallest positive number N such that for each x in the interval (N,+infinity) the value of [tex] f(x)= \frac{x}{x+1} [/tex] is within .01 units of L=1.
    From the definition [tex]|f(x)-L|<\epsilon \mbox{ if } x > N, \epsilon>0, N>0[/tex]. It follows that [tex] |\frac{x}{x+1}-1| < \epsilon \mbox{ if } x > N \mbox{ therefore } |\frac{-1}{x+1}| < \epsilon \mbox{ if } x > N[/tex] Now if I move the minus sign across to epsilon I get [tex] |\frac{1}{x+1}| > -\epsilon \mbox{ if } x > N[/tex]. But epsilon is positive and so is x and so is N yet I will get a negative answer for N according to the last statement. Can anyone clarify the argument needed to get [tex] |\frac{-1}{x+1}| = |\frac{1}{x+1}|[/tex] so that N can remain positive? Thanks
     
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  3. Sep 15, 2009 #2
    i didn't read all that (have to catcha bus) but can't you just go x/(x+1) = 1-0.01 and solve for x?
     
  4. Sep 15, 2009 #3
    Yes, you can just solve for x as you say.
     
  5. Sep 15, 2009 #4

    Hurkyl

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    How'd you do that?

    P.S. use \left| and \right| to make big bars.
     
  6. Sep 15, 2009 #5

    HallsofIvy

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    No, since that is an absolute value, it is larger that negative epsilon for all n.
    [itex]|\frac{-1}{x+1}|= |\frac{1}{x+1}|< \epsilon[/itex] says that [itex]-\epsilon< \frac{1}{x+1}< \epsilon[/itex]

     
  7. Sep 15, 2009 #6
    Hurkyl, I used a numerical example -5< 4 therefore 5 > -4. Thanks for pointing out my error.Thanks all for the replies.
     
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