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Proving a function is not continuous using epsilon and delta definition of a limit

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that f(x) = 1/(x^2) is not continuous at x = 0 using the epsilon and delta definition of a limit


    2. Relevant equations

    definition of discontinuity
    There exist epsilon > 0 such that for all delta > 0 there is an x such that |x-0| < delta but |1/(x^2)| >= Epsilon


    3. The attempt at a solution

    I don't know how to go from here. Like how do i prove it? I don't understand!
     
  2. jcsd
  3. Oct 25, 2011 #2
    Re: Proving a function is not continuous using epsilon and delta definition of a limi

    Start by assuming that [itex]|x| < \delta[/itex]. Consider how this piece of information affects your function [itex]1/x^2[/itex]. Can you derive an [itex]\epsilon[/itex] that is a function of delta such that [itex]|1/x^2| \geq \epsilon[/itex]?
     
  4. Oct 25, 2011 #3

    Deveno

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    Re: Proving a function is not continuous using epsilon and delta definition of a limi

    now, if 1/x2 IS continuous at 0,

    [tex]\lim_{x \to 0} f(x) = f(0)[/tex] which is a problem, because f(0) doesn't exist.

    but let's go even further, let's prove that there isn't ANY real number we can pick for f(0) that will make 1/x2 continuous at 0.

    now, for this limit L to exist, we need to be able to find a δ > 0, so that:

    0 < |x| < δ implies |1/x2 - L| < ε.

    can L be 0? choose ε = 1. that would mean that we could find δ > 0 so that 0 < |x| < δ implies |1/x2| < 1. here, we have another problem.

    |1/x2| = (1/|x|)2, so if 0 < |x| < δ, (1/|x|)2 > δ2. so no δ less than 1 will work. but if we choose δ ≥ 1, then (0,1) is a subinterval of (0,δ), and for 0 < |x| < 1, |1/x2| > 1.

    since for the particular ε = 1, we can't find a δ, L can't be 0. so L > 0 (since 1/x2) (if L < 0, then on the interval (0,1], 1/x2 would have to take on the value 0 (by the intermediate value theorem) but if x ≠ 0, 1/x2 > 0).

    so what happens if δ < 1/(√(2L))?

    then 0 < |x| < 1/(√(2L)) means that |1/x2 - L| ≥ |1/x2| - |L|

    = 1/|x|2 - L > 1/(1/(√(2L))2) - L = 2L - L = L

    now if we choose 0 < ε < L (which we can, since L > 0, for example ε = L/2 would work fine), we have the same problem as before, no δ < 1/(√(2L)) will work, and any larger choice for δ will lead to |x| < 1/(√(2L)) for some x as well (larger δ's don't help).

    so there is always some ε > 0 that doesn't have a δ, no matter what we choose for L. so such an L doesn't exist, we cannot define f(0) to be any real number in such a way as to make f(x) = 1/x2 continuous.
     
  5. Sep 3, 2012 #4
    Re: Proving a function is not continuous using epsilon and delta definition of a limi

    Hello - where did the 1/sqrt(2L) come from? thank you.
     
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