Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The epsilon-delta definition

  1. May 18, 2015 #1
    I'm trying to wrap my head around the epsilon-delta definition.
    "Let ##f## be a function defined on an interval that contains ##a##, except possibly at ##a##. We say that:
    $$\lim_{x →a} f(x) = L$$
    If for every number ##\epsilon > 0## there is some number ##\delta > 0## such that:
    ##|f(x) - L| < \epsilon## whenever ##0 < |x - a| < \delta##"
    Why aren't we restricting ##|f(x) - L|## to be nonzero?
     
  2. jcsd
  3. May 18, 2015 #2

    pasmith

    User Avatar
    Homework Helper

    [itex]|x - a|[/itex] must be bounded away from zero because the concept of a limit deals with what happens to [itex]f[/itex] near to, but not at, [itex]a[/itex]. It might happen that [itex]f(x) = L[/itex] for values of [itex]x[/itex] near to, but not at, [itex]a[/itex]; this is not a problem.

    For example, consider [itex]f: (-1, 0) \cup (0,1) \to \mathbb{R} : x \mapsto 1[/itex]. Then [tex]\lim_{x \to 0} f(x) = 1[/tex] since for any [itex]\epsilon > 0[/itex], if [itex]0 < |x| < 1[/itex] then [itex]|f(x) - 1| = |1 - 1| = 0 < \epsilon[/itex].
     
  4. May 18, 2015 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    For example, if [itex] f(x) [/itex] is the constant function [itex] f(x) = 2 [/itex] whose graph is a horizontal line, we want [itex] \lim_{x \rightarrow 3 } f(x) [/itex] to exist and be equal to [itex] 2 [/itex].
     
  5. May 18, 2015 #4

    Svein

    User Avatar
    Science Advisor

    Why should we? If |f(x)-L| = 0, it is obviously less than ε.
     
  6. May 18, 2015 #5
    So while ##|x - a|## must be nonzero, ##|f(x) - L|## could be zero?
     
  7. May 18, 2015 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The point is that the function value at a does not affect the limit. The limit relates to points close to a, but not a itself.

    But there's no reason the function value can't be equal to the limit.
     
  8. May 19, 2015 #7
    Take for example ##\lim_{x→1} 2x + 3 = 5##.
    We want to prove that for every ##\epsilon > 0## there is some ##\delta > 0## such that:
    ##|(2x + 3) - 5| < \epsilon## whenever ##0 < |x - 1| < \delta##
    This seems to be the general approach taken in most texts/websites:
    ##|2x - 2| = 2|x - 1| < \epsilon## implying that ##|x - 1| < \frac{\epsilon}{2}## which "looks a lot like" ##|x - 1| < \delta##, so we set ##\delta## equal to ##\frac{\epsilon}{2}##.
    Next, we plug in ##\delta = \frac{\epsilon}{2}## into the inequality ##0 < |x - 1| < \delta## and multiply throughout by 2. We end up with ##0 < |2x - 1| < \epsilon## which, again, "looks a lot like" the first inequality, which confirms the fact that our guess was right.
    Isn't that last step redundant? Haven't we already established that ##\epsilon = 2 \delta## by manipulating the first inequality? Isn't this just doing the same exact thing, except this time it's the other way around?
    Also, while the inequalities look the same, there's still a slight difference. We end up with ##0 < |f(x) - L| < \epsilon##. Doesn't this contradict ##|f(x) - L| < \epsilon## (to some extent) since the latter places no restriction on ##|f(x) - L|## being zero?
    And finally, is there another way to look at it other than "this inequality looks an awful lot like the other inequality, therefore ##\delta = \frac{\epsilon}{2}##"? I can't see why ##\delta## and ##\frac{\epsilon}{2}## can't be different despite both being larger than ##|x - 1|##.
     
    Last edited: May 19, 2015
  9. May 19, 2015 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I just want to address this one point. I ask you to find a number ##x##, such that ##|x| < 1##. You suggest ##x = 0.5##. I then say "no". That ##x## satisfies:

    ##0 < |x| < 1## and, therefore, doesn't satisfy ##|x| < 1##

    Note that:

    ##0 < |f(x) - L| < \epsilon \ \ \Rightarrow \ \ |f(x) - L| < \epsilon##

    So, there is no contradiction there.
     
  10. May 19, 2015 #9

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This might help. Rewrite the definition of limit (equivalently) as follows:

    ##\forall \ \epsilon > 0, \ \exists \ \delta > 0## such that ## |x-a| < \delta## and ##x \ne a \ \Rightarrow \ |f(x) - L| < \epsilon##

    Does that clarify things at all?

    All we are doing in both cases is excluding the point ##a## from consideration.
     
    Last edited: May 19, 2015
  11. May 19, 2015 #10

    Svein

    User Avatar
    Science Advisor

    No. We do not take the limit of an equation. We take the limit of a function. So, if you want an example, take [itex]\lim_{x\rightarrow 1}2x+3 [/itex]. First, choose ε: Choose ε=0.1. Then either start calculating or make a guess: Let δ=0.01 and see what happens. On one side: 2*1.001 + 3 = 5.002. On the other side: 2*0.999 + 3 = 4.998. From these two results, we assume that the limit is 5, since |5.002-5| = 0.002 which is less than ε and |5-4.998| = 0.002 which is also less than ε. Now you can either create a formula for δ or you can choose a series of ε's and continue guessing δ's until you have a good idea of how to choose them.

    Now, this may seem obvious, but let me give you another example: Let y = floor(x) (the integer part of x). This function is continuous at all points except when x is an integer (and there it is "continuous from above").
     
  12. May 21, 2015 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Surely this is not what you meant to say!? 1/2 satisfies both ##0< |x|< 1## and ##|x|< 1##.
    Indeed, any number that satisfies ##0< |x|< 1## must also satisfy ##|x|< 1## because ##0< |x|< 1## requires that x satisfy both ## 0< |x|## and ##|x|< 1##.

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The epsilon-delta definition
Loading...