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Homework Help: Epsilon-delta limit definition trouble

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Guess the limit and use the [itex]\epsilon[/itex]-[itex]\delta[/itex] definition to prove that your guess is correct.

    [itex]\lim_{x \to 9}\frac{x+1}{x^2+1}[/itex]

    2. The attempt at a solution

    Guess limit to be [itex]\frac{10}{82}=\frac{5}{41}[/itex]


    [itex]|\frac{x+1}{x^2+1}-\frac{5}{41}| = |\frac{(x-9)(5x+4)}{41(x^2+1)}|[/itex]

    Restrict attention to [itex]|x-9|<1[/itex]

    Therefore: [itex]|\frac{(x-9)(5x+4)}{41(x^2+1)}|<|\frac{54|x-9|}{4141}|<\epsilon[/itex]

    By taking [itex]\delta=min(\frac{4141\epsilon}{54},1)[/itex] we get that:

    [itex]|\frac{x+1}{x^2+1}-\frac{5}{41}| < \epsilon[/itex] whenever [itex]0<|x-9|<\delta[/itex]

    That's my working, but I think I've made a mistake, as when I check my work by using [itex]\epsilon=0.01[/itex] my [itex]\delta[/itex] does not satisfy. This is because I get [itex]0<|x-9|<0.76686 \Rightarrow 8.23314<x<9.76686[/itex]. But then I take say 9.76 and it doesn't hold as I get [itex]|\frac{9.76+1}{9.76^2+1}-\frac{5}{41}| = 0.010168[/itex] which is greater than my [itex]\epsilon[/itex] of 0.01.

    Can anyone help me out with where I went wrong? Thanks!!
    Last edited: Aug 14, 2011
  2. jcsd
  3. Aug 14, 2011 #2


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    41(11)= 451, not 4141.
  4. Aug 14, 2011 #3
    Hi, thanks for reply!

    Sorry, but where is the calculation 41 * 11? I can only think of 41*(10^2+1) :(
  5. Aug 15, 2011 #4
    Found my error. Was I subbed in the largest value of the restricted x range in the denominator, which makes the expression smaller, instead of subbing in the smallest value of the restricted x range (x = 8) into the denominator

    So it becomes: [itex]|\frac{(x-9)(5x+4)}{41(x^2+1)}|<|\frac{54|x-9|}{2665}|<\epsilon[/itex]
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