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

Formal Definition Of A Limit

  1. May 20, 2012 #1
    I am reading about the formal definition of a limit, and its corresponding proof, and there is one thing that I don't quite understand, yet. It says that delta depends on epsilon, but what I wonder is why is it not the other way around. Indeed, why does delta have dependency on epsilon?
     
  2. jcsd
  3. May 20, 2012 #2
    It's basically saying,

    "For every epsilon, there is a delta such that ..."

    Try wrapping your head around how this definition works first, it should be more intuitive then.
     
  4. May 22, 2012 #3

    thrill3rnit3

    User Avatar
    Gold Member

    You want that delta to work for any choice of ε > 0.
     
  5. May 22, 2012 #4

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    To prove the limit exists, if somebody gives you a value of ##\epsilon## (any value they like, so long as ##\epsilon > 0##) you have to find a ##\delta## that satisfies the definition.

    The definition of a limit says something about how the function behaves at a single point, not about how it behaves at every point in an interval. So asking "what happens in an interval of size ##\delta## around the point" and trying to find a value of ##\epsilon## might not make any sense.

    For example think about the function ##f(x) = 1/x## if ##x \ne 0## and ##f(x) = 0## when ##x = 0##. This is continuous at every point except ##x = 0##. But if you take ##x = a## and ##\delta > a## you can't say anything about ##\epsilon## in the interval ##|x-a| < \delta##, which includes points on either side of ##x = 0##. Of course to prove ##f(x)## is continuous at ##x = a##, you use values of ##\delta## that are small enough so the interval does NOT include ##x=0##.
    Work out what the definition means, for some simple functions like ##f(x) = 2x##. For most functions, the smaller the value of ##\epsilon##, the smaller you have to make ##\delta##.
     
  6. May 22, 2012 #5
    In order to constrain the output by a given amount; you must constrain the input by an amount that depends on the contraint on the output.
     
  7. May 22, 2012 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    We want the value of the function to get arbitrarily close to the limit. If we denote the limit by y, this means that for all positive numbers ε, the function must at some point in its domain have a value in the interval (y-ε,y+ε). This is why the definition starts with "For all ε>0".

    Now consider the function f defined by f(x)=sin(1/x) for all x>0. (I edited this sentence after micromass' correction below).

    200px-Topologist%27s_sine_curve.svg.png

    If we only require that the function get arbitrarily close to a limit, then every number in the interval [-1,1] would be a limit of this function at 0. For example, when x approaches 0, f(x) gets arbitrarily close to 1. But we don't want a definition that makes every number in [-1,1] a limit of this function at 0. We want a definition that ensures that this function doesn't have a limit at 0. So we require not only that the function comes close to the limit, but also that it stays close to the limit.

    To be more precise, we say that y is a limit of f at b, or equivalently, a limit of f(x) as x goes to b, if there's an open interval around b such that all the values of f in that interval are in the interval (y-ε,y+ε). In other words, we require that there's a δ>0 such that for all x in (b-δ,b+δ), f(x) is in (y-ε,y+ε).

    If you understand this, it should be pretty obvious that given a small ε, you have to choose a small δ to ensure that the last requirement is satisfied. This is why δ depends on ε.
     
    Last edited: May 22, 2012
  8. May 22, 2012 #7

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    That function plotted isn't sin(x)/x :confused:
    Did you rather mean sin(1/x)?? That would make sense...
     
  9. May 22, 2012 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    No, that's the entire point. You don't want that delta to work for any epsilon. You want a delta to work for each epsilon.
     
  10. May 22, 2012 #9

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Oops. Yes, of course. I will edit my post. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Formal Definition Of A Limit
Loading...