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 limits.

  1. May 16, 2008 #1
    I can't get my head around the epsilon-delta definition of a limit. Unfortunately I don't have a teacher to ask (I'm teaching this to myself as a self interest) so this forum is my last resort -- google hasn't been kind to me.

    From what I've seen, I don't really understand how the definition means much of anything (visual examples included.) All it seems to say is "there is an unspecified value which is greater than the difference between a function and a given value L, where that difference is greater than zero" The problem is, I don't see any need for L or f(x) to be anywhere near one another.

    For example: f(5)==>25 ;

    limx=>6 f(x) = 20

    -1000 < |f(5) - 20| < 1000

    And

    0 < |5-6| < 2

    That is, I see no specific reason to even bother putting in the correct values. How does the definition define anything if we require prior knowledge of what our limits and x values are? Furthermore, how are epsilon and delta even related? The visual explanations make even less sense -- why can't we choose two given values of f(x) and L to which the difference is large, yet select a large enough epsilon so that the centre of f(x) and L lies between the bands of epsilon.

    I'm terribly confused as everyone can probably tell... sorry.
     
  2. jcsd
  3. May 16, 2008 #2
    All it is meant to do is provide a rigorous justification for doing limits. I mean, it's on the one hand just some silly formalism, and on the other hand it does make rigorous proofs more straightforward and feasible.

    The idea is that, if a limit exists, there is an interval (say, of width delta) around the point in which the function is close (say, within a small difference epsilon) to some value, lim f(x). Doing delta-epsilon proofs is just a rigorous way of showing that this interval exists, and thus the limit exists.

    It's a bit awkward, but go ahead and give it a shot. Don't beat yourself up over it.
     
  4. May 16, 2008 #3

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    imagine a function which is monotonically decreasing, and takes on all positive values less than 1, as x approaches 4.

    can you see why, even prove , its limit is zero as x goes to 4?
     
  5. May 16, 2008 #4
    I can't help but get a tingly feeling on my tongue that there's something very profound behind this definition. Unfortunately I'll have to beat myself up over it for a few months before I give up, that's how it always is. =(
     
  6. May 16, 2008 #5
    ALL positive values of less than 1 as x goes to 4? I suppose the smallest positive value less than one is greater than zero by the smallest possible amount, leaving the only value "less than" to be zero itself (that is to say, since "A" is the smallest possible value greater than zero, and since f(x) > f(x+B) (where B is any positive number, since of course the function is always decreasing.) Thus when f(x)==>A, then f(x+B)<f(x), and since the only value less than A is zero itself...


    Errr... I think that makes sense.
     
    Last edited: May 16, 2008
  7. May 17, 2008 #6

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Actually, limits are a very intuitive concept. I think you (and any starting mathematician, even) would agree with me that
    by a statement like [tex]\lim_{x \to a} f(x) = L[/tex] we mean that the closer x gets to a, the closer the function value f(x) gets to L.​
    In other words: if we look at the function value in a point close enough to a, we are certain to find something which is close enough to L. The epsilon-delta stuff is just a way to rigorously define what we mean by "close enough".

    It works as follows:
    • Let [tex]\epsilon > 0[/tex].
    • Then there exists a [tex]\delta > 0[/tex] such that: if [tex]|x - a| < \delta[/tex], then [tex]|f(x) - L| < \epsilon[/tex].
    The first point says, you can choose epsilon. This epsilon will be the maximum allowed difference between the function value and the limit value. You can make it as small or as large as you want. If you keep making it smaller and smaller, the function value will be closer and closer to the limit. Note that it does explicitly say that epsilon is positive: the statement does not say that there will be a point where epsilon is actually zero (that is, the function actually reaches the limit).
    Then the second point asserts the existence of a neighborhood of size delta around a for which you will find the function value within epsilon of L. That is, whatever value of x in the interval [itex](a - \delta, a + \delta)[/itex] I plug into f, I am sure that the function value will be in the interval [itex](L - \epsilon, L + \epsilon)[/itex]. Of course, the smaller the epsilon you give me (the more stringent your definition of what you call "close enough") the smaller my delta will have to be (the closer you will have to be to a to actually come "close enough").
     
  8. May 17, 2008 #7

    Borek

    User Avatar

    Staff: Mentor

    Definition is not a recipe for limit calculation. However, it tells you how to check if the number you suspect to be the limit is the limit - or not.
     
  9. May 17, 2008 #8
    Ohhhh, thanks CompuChip, I was expecting there to be something that implied "[tex]\lim_{\epsilon \to 0}[/tex]" (that is, something which forced epsilon to squash f(x) and L together.) I have to wonder, though, isn't that definition somewhat useless for a function which oscillates extremely rapidly near the limit; for example, sin(1/[tex]x^2[/tex]), where x is very small, say 2E(-10000)?

    Definition is not a recipe for limit calculation. However, it tells you how to check if the number you suspect to be the limit is the limit - or not.

    Amazing that no sites wanted to mention these things. I suppose this is a testimate to text books being much more reliable than internet sources. =P
     
  10. May 17, 2008 #9

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    No, in fact it is extremely useful because it allows you to prove that
    [tex]\lim_{x \to 0} \sin(1 / x^2)[/tex]
    does not exist (which you of course already suspected by a reasoning like: "if x goes to zero, 1/x^2 goes to infinity, and the sine function keeps oscillating there, so probably there is no limit"). For suppose you claim that the limit is [itex]-1 \le L \le 1[/itex], then I can show that there is an epsilon such that for all delta, there is an x closer than delta to zero for which sin(1/x^2) will be more than epsilon from L. Which by the very definition says: "the limit is not L".
     
    Last edited: May 17, 2008
  11. May 17, 2008 #10

    Borek

    User Avatar

    Staff: Mentor

    I am afraid text books will be no better here. Many math books leave lots of things untold - as these are obvious. Fact that these things are obvious only for mathematicians, doesn't matter - at least to them :wink:
     
  12. May 18, 2008 #11

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    I forgot about this part:

    Note that there is no such thing explicitly. But of course, they idea is that you will make epsilon smaller and smaller. You are allowed to choose [itex]\epsilon = 10^{10^{100}}[/itex], and the statement should still be true (and in fact, chances are it is, unless you have chosen the suspected limit really badly or you have a strange function). The point is, however, that it holds for all [itex]\epsilon > 0[/itex], no matter how small. So if there is a number for which it is not true, whether it be very large or very small, then the definition is not satisfied. In fact, as I already stated implicitly in my last post, the opposite of
    [tex]\lim_{x \to a} f(x) = L[/tex], meaning: for each [itex]\epsilon > 0[/itex] there is a [itex]\delta = \delta(\epsilon) > 0 [/itex] such that for all x satisfying [itex]|x - a| < \delta[/itex], [itex]|f(x) - L| < \epsilon[/itex] ​
    is
    [tex]\lim_{x \to a} f(x) \neq L[/tex], meaning: there is an [itex]\epsilon > 0[/itex] (finding just one suffices) such that for all [itex]\delta = \delta(\epsilon) > 0 [/itex] there is an x satisfying [itex]|x - a| < \delta[/itex] for which [itex]|f(x) - L| \ge \epsilon[/itex] ​
    Which is: there is a distance [itex]\epsilon[/itex] for which I cannot get f(x) closer than [itex]\epsilon[/itex] to the supposed limit L, whence it cannot be the limit.
     
  13. May 18, 2008 #12
    I meant when x is very small, but not EXACTLY zero. For examle, [tex]\lim_{x \to 10^{-10000}} \sin(1/x^2)[/tex] The required value for epsilon and delta is so incredibly small that it seems rather difficult to "pick a value that 'seems close enough," it would be very easy to accidently pick a value too large or too small. I suppose I was hoping the formal definition would introduce something less... ambiguous (though I suppose it needs to be based on the nature of a limit.) It is, as you say, very useful, just not the solution to all your limit needs. Anyway, many thanks for the advice, everyone.
     
  14. May 19, 2008 #13

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    I don't really see what your problem with that definition is. There is no "required value" of epsilon, you can choose it, as I tried to explain. For any given epsilon, finding the right value of delta can indeed be a tough job, in fact, that's the hardest part about epsilon-delta proofs.

    Also, the "problem" with the function you introduced only occurs at x = 0, where the function indeed has no limit (as you can actually prove from the definition). You can also prove from the definition that [tex]
    \lim_{x \to 10^{-10000}} \sin(1/x^2) = \left. \sin(1 / x^2) \right|_{x = 10^{-10000}}
    [/tex]
    (or you can prove first that if f is continuous in a, then [itex]\lim_{x \to a} f(x) = f(a)[/itex]). It is not any easier or more difficult than proving that [tex]\lim_{x \to 1} \sin(1 / x^2) = \sin(1)[/tex]. In general in such proofs, you start by writing down "Let [itex]\epsilon > 0[/itex]" and then proceed to define delta in terms of epsilon, e.g. "Choose [itex]\delta = \sqrt{\epsilon} / 4[/itex] and let x be such that [itex]|x - a| < \delta[/itex]." Then you go and estimate the difference f(x) - L in a series of inequalities to show that it is smaller than epsilon. Whether a is 0 or infinity, or something in between, is irrelevant for that structure of the proof and does not increase or decrease the level of difficulty.

    Can you explain what it is that you find ambiguous about the definition?
     
    Last edited: May 19, 2008
  15. May 19, 2008 #14
    The simpler, the better...

    Find the limit of f(x) = x when x goes to infinity, I choose you epsilon = 17 and 3506, can you give me a delta such that the conditions satisfied?

    What about f(x) = 1/x? I give you epsilon = 1849 and 0.02? Any deltas?

    And lastly, f(x) = cos(x), I choose epsilon = 0.5...
     
  16. May 20, 2008 #15
    That the definition doesn't say what value epsilon or delta should be. By the second part we say [itex]0 < |x - a| < \delta[/itex]. That's saying that no matter how small we make delta, the difference of x and a will always be smaller -- but not fully zero? It's SMALLER than ANY number you could EVER choose, but NOT zero? What the heck is the reciprocal of such a small number? BIGGER than any number you could choose? I understand what the definition is saying, it just seems weird.
     
  17. May 20, 2008 #16

    Borek

    User Avatar

    Staff: Mentor

    Because it doesn't matter what values they have, as long as they fulfill the condition.
     
  18. May 20, 2008 #17

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    That's because epsilon can be anything (as I explained), and delta depends on epsilon and the function itself. The definition works for any function, exactly because it doesn't prescribe delta.

    Sorry, I missed the part where it says that |x - a| cannot be zero. The definition says: if |x - a| < delta, then |f(x) - L| < epsilon. In particular, you can do it for x = a, if the function is defined there (i.e. for a continuous function). But for functions like 1/x you cannot make |x - 0| equal to 0 because that would force you to choose x = 0 but the function is not defined there.

    In the phrase "such a small number", which number are you talking about? Epsilon? Delta?
    And the reciprocal of a small number is indeed a large number, but what does that have to do with it?

    Actually, once you understand what the definition is trying to express, I think it's a completely logical way to do it, and one of the few ways that expresses the idea of a limit without making use of that concept. So if you don't mind me being very blunt, I don't think you fully understand it yet :wink:
     
  19. May 20, 2008 #18
  20. May 20, 2008 #19
    Yeah that's right you don't consider x = a in the definition. The limit can exist even if the function is not defined at that point (which would not be true if you needed to include x=a in the definition). For example [tex]f(x) = \frac{x^2-1}{x - 1}[/tex] is not defined at x=1 but [tex]\lim_{x\rightarrow 1} f(x) = 2[/tex].
     
  21. May 20, 2008 #20

    lurflurf

    User Avatar
    Homework Helper

    Infact the values of f only need be considered when 0<|x-delta|<epsilon for any epsilon
    no only can the function be undifined elsewhere it may be differntly defined the limit checks if the function is almost constant in any small neighborhood of the value
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Epsilon delta limits.
  1. Epsilon Delta Limit. (Replies: 5)

Loading...