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Homework Help: Prove the LimitI still don't understand these

  1. Sep 18, 2009 #1
    Alrighty-then :smile:

    I am really bad ay these I guess. I keep trying to reason these out, but i get stuck in the same spot every time.

    Here we go...wheeeee!



    So I start by working 'backwards' to find a suitable [itex]\delta[/itex]







    Here is where I keep getting stuck. I am really bad at generalizing the inequality to find out which 'side' is closer to xo

    I have assumed that [itex]0<\epsilon<1[/itex] for now.

    Using that assumption and looking at each side of the inequality we have:

    a) [tex]\frac{1}{1-\epsilon}-1[/tex]


    b) [tex]\frac{1}{1+\epsilon}-1[/tex]

    But I still cannot see which of those to use. If [itex]\epsilon[/itex] is close to 0, then (a) is also close to zero and (b) is close to zero but negative.

    But if [itex]\epsilon[/itex] is close to 1, then (a) becomes very large and (b) is close to 1/2.

    Can someone shed some light on this?

  2. jcsd
  3. Sep 18, 2009 #2
    I think this is incorrect.

    [tex]a < b \not\Rightarrow \frac{1}{a} < \frac{1}{b}[/tex]

    If you correct this, I think things may work more smoothly.

    In general if a < 0 < b and a < X < b then |X| < min(|a|, b). Frequently it is possible to determine which of the two expressions is less.

  4. Sep 18, 2009 #3

    Sorry Euclidus, I am a little tired. What are you referring to as 'a' and 'b' ? :redface:

    EDIT: Oh, I see. a=1-e and b=1+e
  5. Sep 18, 2009 #4
    However this
    does not make sense to me.

    Given -1 < 3 < 5

    |3| is not less than |-1|
  6. Sep 18, 2009 #5
    I see that I had the inequality reversed, but that still doesn't help me to determine which one is smaller.

    Any ideas anyone?

    EDIT: After poking around online a little I have a question about this: Do I actually have to determine which is smaller?

    Or is it enough to just say that delta = MIN{a,b} ?
    Last edited: Sep 18, 2009
  7. Sep 18, 2009 #6
    Sorry, I was trying to help get to a good delta value.

    What I was trying to get across was that if you need a < X < b (with a < 0 < b) then choosing |X| < min(|a|, b) will guarantee that. The implication is:

    If a < 0 < b and |X| < min(|a|, b) then a < X < b.

    In relationship to the problem at hand:

    If |x - 1| < min(e/(1-e), e/(1+e)) then -e/(1+e) < x - 1 < e/(1-e).

    But e/(1+e) < e/(1-e) for 0 < e < 1.

    I hope this is helpful.

  8. Sep 18, 2009 #7
    Okay. So you were not saying that for all a<X<b => |X| < min{|a|,b}.

    You were saying choose X such that it meets the above conditions?

    As for this problem, I guess that I am having trouble seeing how to show that this "But e/(1+e) < e/(1-e) for 0 < e < 1" is true for
    all 0 < e <1
  9. Sep 19, 2009 #8
    Although your algebraic approach works, I think the epsilon-chasing overcomplicates things a bit. It's useful to learn how to use min and max of numbers to satisfy multiple inequalities, but this problem lends itself to thinking geometrically.

    Note that if you work with the Euclidean distance, then the condition |f(x) - L| < e becomes |(1-x)/x| = |x-1| * (1/|x|) < e. Now you only need to find an upper bound for 1/|x| and ensure that x does not equal 0. We still have |x-1| < d at our disposal, so that we can require x to be within delta of 1 or equivalently, we can choose delta so that x is in (1-d, 1+d). Now can you see how to choose delta so that we can actually restrict x to be positive? This will also give an upper bound for 1/|x|, and then the problem is basically solved (well okay you still have to use the min function, but if you got the gist of this approach, the details for choosing delta are not hard to formalize).
  10. Sep 19, 2009 #9
    I just don't see why this is so hard for me to get? My biggest problem is this line (in order to choose delta):


    For 0 < [itex]\epsilon[/itex] < 1 , how can I prove that

    [tex]\frac{1}{1+\epsilon}-1 < \frac{1}{1-\epsilon}-1[/tex]


    Because I am not quite convinced that it is. Sorry, I am slow.
  11. Sep 19, 2009 #10
    Cancel -1 from both sides, clear denominators (which we can do since (1-e) and (1+e) are positive for 0 < e <1) and we're left with 1 - e < 1 + e which is e > 0.
  12. Sep 19, 2009 #11
    Okay. I can live with that.

    Now, I have never actually completed the proof before. We have solved the inequality to tease out a value of delta. I think I am supposed to do something with it formally now.

    I believe that in general, the proof is that given some [itex]\epsilon>0[/itex], there exists some corresponding [itex]\delta[/itex] such that for all 'x'

    [itex]0<|x-x_o|<\delta \Rightarrow |f(x)-L|<\epsilon[/itex]

    So now I need to apply that to this problem:

    Given [itex]\epsilon[/itex], choose [itex]\delta=\frac{1}{1+\epsilon}-1[/itex] so that

    [itex]|x-1|<\frac{1}{1+\epsilon}-1\Rightarrow |\frac{1}{x}-1|<\epsilon[/itex]

    I am just not sure what I do with these two statements now? How do prove (formally) that by choosing delta as such, the inequality holds?

    Do I just 'evaluate' [itex]|x-1|<\frac{1}{1+\epsilon}-1[/itex] for 'x' and then plug the results into [itex]|\frac{1}{x}-1|<\epsilon[/itex]

    I think so. Let's have a go.... back in a minute :smile:

    EDIT: I'm back.

    So now we have




    So now do I just test each of these in the inequality [itex]|\frac{1}{x}-1|<\epsilon[/itex]?
    Last edited: Sep 19, 2009
  13. Sep 20, 2009 #12
    I tried plugging in [itex] 2-\frac{1}{1+\epsilon}\text{ into }|\frac{1}{x}-1|<\epsilon[/itex] but it seems to be just reducing to something silly like [itex]2\epsilon>0[/itex] which doesn't say much.

    Was I supposed to 'plug' [itex]2-\frac{1}{1+\epsilon}[/itex] for 'x' in [itex]|\frac{1}{x}-1|<\epsilon[/itex]?
  14. Sep 20, 2009 #13
    You need to take more care in checking your work. Here's your problem: you chose a negative delta. If you look back earlier in this thread, you'll see that we reached two equivalent series of inequalities:

    [tex]\frac{1}{1+\varepsilon} - 1 < x - 1 < \frac{1}{1-\varepsilon} - 1 \leftrightarrow -\frac{\varepsilon}{1 + \varepsilon} < x - 1 < \frac{\varepsilon}{1-\varepsilon}.[/tex]

    The left hand side of the last chain of inequalities is negative. I'll leave it to you to figure out why you have to pick the smaller of the absolute value of the expressions to the left and right of x - 1 in the last chain of inequalities as your delta.

    This should really finish the problem, barring any mistakes on my end. I suggest you read the very first post I made in this thread. I did not make it so you could bash yourself more over figuring out why your first approach is taking longer than you expected to verify. Your first approach as a whole is not particularly instructive and probably not worth checking the algebraic details.
  15. Sep 20, 2009 #14
    Thanks guys for all of your effort and I am sorry that I just do not understand this stuff. I absolutely do not want to sound ungrateful or unappreciative, because I am neither.

    But my main issue with this thread is that I posed a problem and a particular solution path.

    Perhaps it was not the best solution path, but it was the one that was taught in the text. And though I am absolutely in agreement that one should learn as many different approaches as possible to the same problem, surely one must learn them one at a time, yes?

    So...snipez90: I am sure that your post #8 is a great approach and I will start dissecting it ASAP, for now I would like to complete the problem as stated in post #1.


    With regard to the double inequality

    [itex]\frac{1}{1+\varepsilon} - 1 < x - 1 < \frac{1}{1-\varepsilon} - 1 \leftrightarrow -\frac{\varepsilon}{1 + \varepsilon} < x - 1 < \frac{\varepsilon}{1-\varepsilon}[/itex]

    well, somehow the inequality on the right of the arrow slipped past me.

    I have know idea how we arrived at that. I know that it was in Euclidus's post #6. Now, are the two equivalent? I am sure that in a few minutes I can answer that on my own, BUT the bigger question is WHY did we decide to put it into that form in the first place?
  16. Sep 20, 2009 #15
    Well you kind of answered this yourself here:

    There is no possible way you're going to find a delta that works by guessing (well sometimes you do, but generally this does not work). You always start by determining how close you want f(x) to be to the limit L, and then determine how sufficiently close you need x to be to x_0 so that the degree of closeness between f(x) and L is actually achieved. Thus, we pick an arbitrary [itex]\varepsilon > 0,[/itex] because if the limit really does exist (which it should), then we can make f(x) as close to L = 1 as we like, which is the same thing as saying that we can make|f(x) - 1| < [itex]\varepsilon[/itex] for any [itex]\varepsilon > 0[/itex] (if you don't understand this part geometrically and intuitively, then revisit that before working algebraically).

    Now that we have picked an arbitrary positive epsilon, we need to determine how close x must be to x_0 = 1 so that |f(x) - 1| < [itex]\varepsilon[/itex] holds. Similar to what was just discussed in the above paragraph, making x close to x_0 = 1 means making |x - 1| small. How small? Small enough so that |f(x) - 1| < [itex]\varepsilon[/itex] is satisfied. If we have any hope of doing this, we have to somehow relate the expression f(x) - 1 to x - 1, and the easiest way to do this is to work backwards . Again, we have already prescribed how close we want f(x) and L to be, i.e. we have picked an arbitrary epsilon so that |f(x) - L| < epsilon, so trying to manipulate f(x) -1 to get some info on how close x must be to 1 (how small |x-1| needs to be) makes more sense than doing it the other way around. Now I retrace everything you did in the first post without the algebraic mistake at one step (hopefully you have a better idea of what our goal is):

    [tex]|f(x) - L| < \varepsilon \Leftrightarrow |\frac{1}{x} - 1| <\varepsilon \Leftrightarrow -\varepsilon < \frac{1}{x} - 1 < \varepsilon \Leftrightarrow 1 - \varepsilon < \frac{1}{x} < 1 + \varepsilon.[/tex]

    At this point, there is a slight hitch, because even though we normally care about very small epsilon, epsilon can be much larger than 1, and in this case, [itex]1 - \varepsilon < 0[/itex] which means that flipping the inequality [itex]1 - \varepsilon < \frac{1}{x}[/itex] can lead to a false statement. But for now, we may temporarily require that [itex]0 < \varepsilon < 1[/itex] and further manipulate the last expression. It might be that in the end, we can require x to be sufficiently close to 1 so that any epsilon will work. Note that so far, all of our inequalities are reversible, which means that we can start with the last chain of inequalities and conclude the very first. We can reverse our steps because all we've done is basic algebra. Continuing from above under the temporary assumption [itex]0 < \varepsilon < 1,[/itex] we have

    [tex]1 - \varepsilon < \frac{1}{x} < 1 + \varepsilon \Leftrightarrow \frac{1}{1+\varepsilon} < x < \frac{1}{1 - \varepsilon} \Leftrightarrow \frac{1}{1+\varepsilon} - 1 < x - 1 < \frac{1}{1 - \varepsilon} - 1 \Leftrightarrow -\frac{\varepsilon}{1 + \varepsilon} < x - 1 < \frac{\varepsilon}{1 - \varepsilon}.[/tex]

    Now we know that if x - 1 satisfies the last chain of inequalities and 0 < epsilon < 1, then f(x) will be within epsilon of L (i.e. |f(x) - L| < epsilon). However, we generally would like to have symmetry when talking about how close we want a point x to be to another point y on the real line. For instance, |f(x) - L| < epsilon means L - epsilon < f(x) < L + epsilon, which means that when we let epsilon become smaller and smaller, we are "closing in" on the limit L and requiring that f(x) be in the open interval (L - epsilon, L + epsilon). We have satisfied our goal of writing f(x) - 1 in terms of x - 1, but remember the reason we wanted to do this in the first place is so that we can require |x-1| to be small enough so that our function f will map these x values to f(x) such that |f(x) - 1| < epsilon. But remember requiring |x - 1| to be small enough means that we want |x-1| < [some positive number], similarly to what was done for the degree of closeness between f(x) and L = 1. So let d > 0 be a positive number. How do we choose d so that |x-1| < d implies that f(x) is within epsilon of L? Well we have

    [tex]-\frac{\varepsilon}{1 + \varepsilon} < x - 1 < \frac{\varepsilon}{1 - \varepsilon},[/tex]

    which tells us something about how the distance relationship between x and 1. It tells us that x is between two quantities:

    [tex]1 -\frac{\varepsilon}{1 + \varepsilon} < x < 1 + \frac{\varepsilon}{1 - \varepsilon},[/tex]

    On the other hand, writing out |x-1| < d as 1 - d < x < 1 + d tells that x is between 1 - d and 1 + d, and we note the symmetry discussed above. But

    [tex]1 -\frac{\varepsilon}{1 + \varepsilon} < x < 1 + \frac{\varepsilon}{1 - \varepsilon},[/tex]

    is not very symmetrical. To remedy this, if we choose the smaller of the two expressions:

    [tex]\frac{\varepsilon}{1 + \varepsilon} and \frac{\varepsilon}{1 - \varepsilon},[/tex]

    as our number d, will this work? In general yes, since if we have 1 - A < x < 1 + B for two distinct positive numbers A and B, then if A < B, then 1 - A < 1 + A < 1 + B, so if something is true for 1 - A < x < 1 + B, then it is also true for 1 - A < x < 1 + A, since 1 + A < 1 + B (and we can choose d = A). Similarly you can work out the case when A > B > 0 to find that if something is true for x such that 1 - A < x < 1 + B then it is also true for x with 1 - A < 1 - B < x < 1 + B (and here d = B). By some basic algebra, we find that in this case,

    [tex]\frac{\varepsilon}{1 + \varepsilon} < \frac{\varepsilon}{1 - \varepsilon}[/tex]

    for 0 < epsilon < 1. Now choose d = [itex]\frac{\varepsilon}{1 + \varepsilon}.[/itex]

    By what we just said, it should be the case that

    [tex]1 - \frac{\varepsilon}{1 + \varepsilon} < x < 1 + \frac{\varepsilon}{1 + \varepsilon} < 1 + \frac{\varepsilon}{1 - \varepsilon},[/itex]

    and noting that last chain of inequalities from when we manipulated f(x) - 1 to something involving x - 1, our choice of d should suffice to show that f(x) is within epsilon of 1. Let us check that it does. We have

    [tex]|x-1| < d = \frac{\varepsilon}{1 + \varepsilon} \Rightarrow - d < x - 1 < d \Rightarrow -\frac{\varepsilon}{1 + \varepsilon} < x - 1 < \frac{\varepsilon}{1 + \varepsilon} \Rightarrow 1 - \frac{\varepsilon}{1 + \varepsilon} < x < 1 + \frac{\varepsilon}{1 + \varepsilon} \Rightarrow \frac{1}{\varepsilon + 1} < x < \frac{2\varepsilon + 1}{\varepsilon + 1}.[/tex]

    Note that our choice of d has gotten rid of earlier problems involving taking reciprocals of both sides of an inequality since we restricted our inequality for the sake of symmetry. Note that x is clearly positive, so we may take reciprocals, taking care to flip the inequality sign as well:

    [tex]\frac{1}{\varepsilon + 1} < x < \frac{2\varepsilon + 1}{\varepsilon + 1} \Rightarrow \frac{\varepsilon + 1}{2\varepsilon + 1} < \frac{1}{x} < \varepsilon + 1 \Rightarrow \frac{\varepsilon + 1}{2\varepsilon + 1} - 1 < \frac{1}{x} - 1 < \varepsilon + 1 - 1 \Rightarrow \frac{-\varepsilon}{2\varepsilon + 1} < \frac{1}{x} - 1 < \varepsilon.[/tex]

    We just have to check that

    [tex]\frac{-\varepsilon}{1 + 2\varepsilon} > -\varepsilon \Leftrightarrow \frac{\varepsilon}{1 + 2\varepsilon} < \varepsilon \Leftrightarrow \varepsilon < \varepsilon + 2\varepsilon ^2,[/tex]

    and this last inequality is clearly true. Hence

    [tex]-\varepsilon < \frac{-\varepsilon}{2\varepsilon + 1} < \frac{1}{x} - 1 < \varepsilon[/tex]

    so [itex]-\varepsilon < \frac{1}{x} - 1 < \varepsilon \Rightarrow |\frac{1}{x}-1| < \varepsilon \Rightarrow |f(x) - 1| < \varepsilon,[/itex]

    which is what we wanted.
    Last edited: Sep 20, 2009
  17. Sep 20, 2009 #16
    Hi there snipez 90. I am almost there with you! I think that you may have misinterpreted this question:

    What I meant by that is since the inequality

    [tex] \frac{1}{1+\varepsilon} - 1 < x - 1 < \frac{1}{1 - \varepsilon} - 1 \text{ is EQUIVALENT TO } -\frac{\varepsilon}{1 + \varepsilon} < x - 1 < \frac{\varepsilon}{1 - \varepsilon}[/tex]

    why do we use the 2nd inequality? Why is it advantageous to have it in that form? You see, if you or Euclidus hadn't written it that way, it never would have occurred to me to put it in that form.

    I am just wondering how one knows that one particular form of an expression is better then another? (I am guessing that it just comes down to experience with different problems.)
  18. Sep 20, 2009 #17
    Try reading my last post again where I specifically mentioned the issue of symmetry. There is not some mystical reason to choose one inequality over the other. It's simply a matter of simplifying as much as possible and making it easier to choose/recognize "delta".
  19. Sep 20, 2009 #18
    Ok. But I am still not sure how that answers the question :confused: I have read your post many times. And I think I get it with the exception of that one part.

    I'll just live with it and move on. I am sick of this problem anyway.

    Thanks for all you help! :smile: I really appreciate it!
  20. Sep 20, 2009 #19
    The whole point is simplify. It doesn't make a difference which inequality you use if they are equivalent. Sure, I did not have to write (1/(1+e)) - 1 as one fraction, but for one thing, you might overlook the fact that (1/(1+e)) - 1 is negative and choose it for delta. I emphasize the fact that it doesn't make a difference which inequality you use IF you are careful, but it probably makes a few subsequent calculations much easier as well. I too am sick of this problem, since we are going back and forth about the tiny details in an epsilon-delta proof, which matters little in the grand scheme of things.

    I have a friend who was pretty bad at proving limits that I asked him to check over about a year ago, yet he did exceptionally well in possibly the hardest undergraduate course at our university (honors version of real analysis). The basic epsilon-delta definition and the intuition behind it is useful, whereas proving limits that you already know exist and are intuitively clear is in general not particularly useful.
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