1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    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!

Epsilon-delta proof for a limit of a troublesome function

  1. Mar 28, 2012 #1
    limx→01/(x2-1)=-1

    0<|x|<δ and |((1/(x2-1)) +1)<ε

    Working on the second equation, I eventually got
    |x|< (ε|x-1||x+1|)0.5
    Suppose I restrict |x|<1
    -1<x<1,
    then
    0<x+1<2 → |x+1|<2
    -2<x-1<0 → 0<1-x<2 → |1-x|=|x-1|<2 (Is this step correct? In one of the textbooks, it says that we must always have x>0 which is important to conclude that for example, |x-1|<2 in the context of this question. Unfortunately, I don't see why we must conclude that x>0 because doesn't the modulus get rid of all negative values?)

    So my choice for δ works out to be δ=min{1,2(ε)0.5}

    However when I work backwards the problem arises.
    Suppose I choose δ=1 → 2(ε)0.5 > 1 → ε>0.25

    When I draw the graph, my δ-range of x=0 starts from the asymptote at x=-1 to the other asymptote x=1, and if this is the case, doesn't that mean the range ε about f(x)=-1 is infinite?
     
  2. jcsd
  3. Mar 28, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have$$
    \left |\frac 1 {x^2-1}+1\right |=\left | \frac {1+x^2-1}{x^2-1}\right |= \left |\frac{x^2}{x^2-1}\right |$$which you want to make small. The numerator will be small as ##x\to 0## but if ##x## gets anywhere near ##\pm 1##, that denominator will make the fraction large. So you must keep ##x## away from ##\pm 1##. That is easy since you have ##x\to 0##. So start by restricting ##|x|<\frac 1 2##. Figure out the smallest ##|x^2-1|## can be and overestimate ##\frac{x^2}{|x^2-1|}## by putting that smallest value in the denominator. Then you will be ready to figure out ##\delta## to make it all less than ##\epsilon##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Epsilon-delta proof for a limit of a troublesome function
Loading...