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Homework Help: Another delta, epsilon problem

  1. Jul 24, 2010 #1
    1. The problem statement, all variables and given/known data

    For:

    [tex] \lim_{x \to 1^{-}} \frac {1}{1-x^{2}} = \infty [/tex]

    Find [tex]\; \delta > 0 \;[/tex] such that whenever:

    [tex] 1-\delta<x<1 \;\;[/tex] then [tex] \; \frac {1}{1-x^{2}} > 100 [/tex]

    2. Relevant equations

    [tex] |x-a| < \delta [/tex]

    [tex] |f(x)-L| < \epsilon [/tex]

    3. The attempt at a solution

    So as it is set right now, this:

    [tex] \frac {1}{1-x^{2}} > 100 [/tex]

    when solved for x will give me the upper limit but I am looking for the lower limit so I solve for x starting with this:

    [tex] \frac {1}{1-x^{2}} < -100 [/tex]

    [tex] 1-x^{2} > -10^{-2} [/tex]

    [tex] -x^{2} > -10^{-2}-1 [/tex]

    [tex] x^{2} > 10^{-2}+1 [/tex]

    [tex] x > \sqrt{10^{-2}+1} [/tex]

    and then since:

    [tex] 1-\delta<x<1 = -\delta<x-1<0 [/tex]

    then I will subtract 1 from both sides of the inequality to get:

    [tex] x-1 > \sqrt{10^{-2}+1}-1 [/tex]

    which works out to be:

    [tex] x-1 > .0049 \; or \; \approx 5*10^{-3} [/tex]

    which 5*10^-3 is the correct answer but I am not sure that I did the work correctly. They got an exact answer of:

    [tex] \delta = 1 - \sqrt{.99} [/tex]

    which is actually about .00504, so not my answer exactly

    Someone help with where I went wrong?

    I say that 5*10^-3 is correct because the answer in the back reads like so:

    [tex] \delta = 1 - \sqrt{.99} \; or \; \approx 5*10^{-3} [/tex]

    thanks!
     
    Last edited: Jul 24, 2010
  2. jcsd
  3. Jul 24, 2010 #2

    vela

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    You don't want to do that because...
    You have x>1 but you're trying to find the lower limit of x as you approach from the left.
     
  4. Jul 24, 2010 #3
    So are you saying I should have solved it as:

    [tex] \frac {1}{1-x^{2}} > 100 [/tex]

    instead of swapping it around? I thought that my way was correct because you end with the sign as something less than x and thus that would be the lower limit as it is less than x, is that incorrect logic?
     
  5. Jul 24, 2010 #4

    vela

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    Right, you shouldn't have swapped. You ended up solving the problem for x approaching 1 from the right.
     
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