1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another delta, epsilon problem

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


    [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]

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Right, you shouldn't have swapped. You ended up solving the problem for x approaching 1 from the right.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Another delta epsilon Date
Jacobian of a coordinate system wrt another system Feb 9, 2018
Yet Another Delta-Epsilon Jul 22, 2010
Another Delta-Epsilon Question Jul 19, 2010
Another epsilon-delta proof Oct 25, 2009
Another epsilon-delta proof Oct 17, 2009