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Relationship between division, subtraction, and limits

  1. Sep 24, 2013 #1
    Ok, so I'm not really too good at group theory and that kind of math, so I hope I can explain my question:

    I tried to evaluate [itex]\frac{d}{dx}e^{x}[/itex]:

    [itex] \frac{d}{dx}e^{x} = \frac{e^{x+h}-e^{x}}{h}[/itex], h -> 0
    [itex] = \frac{e^{x}e^{h}-e^{x}}{h}[/itex], h-> 0
    [itex] = e^{x}(\frac{e^{h}-1}{h})[/itex], h-> 0

    So I figured if I could show that: [itex] \frac{e^{h}-1}{h}[/itex] goes to 1 as h goes to 0, then the proof would be done. So I reasoned that [itex]e^{h}[/itex] goes to 1 as h goes to 0, so [itex]e^{h}-1 [/itex] must go to 0 as h goes to 0. But does [itex]e^{h}-1[/itex] "go to h" as h goes to 0?

    It took me a little bit to phrase this mathematically, but I reasoned that if the difference between [itex]e^{h}-1[/itex] and h went to 0, as h went to 0, then their ratio would get closer and closer to 1, as h went to 0, and the proof would be done. So I tried it:

    [itex]e^{h} - 1 - h = 1 - 1 - 0 = 0 [/itex], as h goes to 0.

    So I figure that this means that [itex]\frac{e^h-1}{h} = 1[/itex], as h goes to 0. I don't know whether or not this is true, but I assume that it is because I don't see how else the derivative would end up as [itex]e^{x}[/itex]. So this brings me to my question:

    So the difficult part of my problem began with two functions, which I will assume to be elements of some ring or defined on some ring (I'm not sure what ring theorists would say here), they are: [itex]e^{h}-1[/itex], and h.

    I encountered a problem where the multiplicative inverse of h was multiplied to [itex]e^{h}-1[/itex]. I did not know how to take the limit of this as h went to 0.

    I translated this into a problem where the additive inverse of h was added to [itex]e^{h}-1[/itex]. I DID know how to take the limit of this.

    Taking the limit as h goes to 0 of this additive version of the problem, I discovered it was equal to 0. This (as far as I know) just happens to be the additive identity.

    Using this information, I concluded that the limit as h goes to 0 of the multiplicative version must be 1. This (as far as I know) just happens to be the multiplicative identity.

    ... This is where I'm not sure how to phrase my question ... This was a nifty little trick to solve this one problem, but there seems to be a lot of structure and patterns here, and, not knowing very much group theory, I don't know what to make of it.
    1) Is there some deeper thing going on here that group theory explains?
    2) I.e. is this some kind of relationship between the addition part and multiplication part of a single RING?
    3) Or is this like a "homomorphism" between a GROUP where the operation is addition, and another GROUP where the operation is multiplication?
    4) Does this apply more generally to other groups/rings/fields, etc?
    5) Is there some more general procedure to translate difficult problems involving multiplication into easier ones involving addition? How would I know when something like this can be applied?
    6) Does the fact that I'm taking limits have anything to do with it?
    7) Does the fact that the limit points are the additive and multiplicative identities have anything to do with it?
    8) Does the continuity of the real numbers and the existence of limits have anything to do with it?

    Thanks in advance for any help in understanding this
  2. jcsd
  3. Sep 24, 2013 #2


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    The type of limit (of the form ##0/0##) that you are taking is called an indeterminate form and these are often tricky to evaluate. I'm afraid that you've gotten the correct answer using somewhat specious reasoning.

    The problem with an indeterminate form is that its value strongly depends on the way that the numerator and denominator go to their individual limits. This is explained on the wiki, but deserves a separate demonstration here. The limit involved here is indeed

    $$ \lim_{h\rightarrow 0} \frac{e^h -1}{h} = \lim_{h\rightarrow 0} \frac{h}{h} = 1.$$

    However, you concluded that

    $$ \lim_{h\rightarrow 0} (e^h -1) =\lim_{h\rightarrow 0} h $$

    because you found that

    $$ \lim_{h\rightarrow 0} (e^h -1 -h ) =0.$$

    Unfortunately, this is not enough, because one can see that

    $$ \lim_{h\rightarrow 0} (e^h -1 ) = 0 ~~~~\mathrm{and} ~~~~ \lim_{h\rightarrow 0} h =0.$$

    independently. So we are really just adding 0 to 0 and haven't proven anything.

    The further problem with the indeterminate form is that

    $$ \lim_{h\rightarrow 0} (e^{h^{1/2}} -1 ) = 0$$

    as well, so we also have

    $$ \lim_{h\rightarrow 0} (e^{h^{1/2}} -1 -h ) = 0.$$


    $$ \lim_{h\rightarrow 0} \frac{e^{h^{1/2}} -1}{h} = \lim_{h\rightarrow 0} \frac{h^{1/2}}{h} = \lim_{h\rightarrow 0} \frac{1}{h^{1/2}} = \infty$$

    is not defined. Similarly,

    $$ \lim_{h\rightarrow 0} \frac{e^{h^2} -1}{h} = \lim_{h\rightarrow 0} \frac{h^2}{h} = \lim_{h\rightarrow 0} h=0.$$

    We are led to conclude that the rate at which the numerator and denominator go to zero are very important in taking a limit of this type, so we must actually use a better method to evaluate the limit in the numerator. One valid method to define the limit (without using calculus, since the calculus result is what we wanted to prove) is to use the limit definition of the exponential function

    $$ e^h = \lim_{n\rightarrow \infty} \left( 1 + \frac{h}{n} \right)^n.$$

    We can use the binomial formula to compute the terms in this expression and take the limit ##h\rightarrow 0## to obtain the correct result.
  4. Sep 25, 2013 #3

    D H

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    A perhaps more intuitive method is to use the power series definition of the exponential function, ## e^h = \sum_{n=0}^{\infty} \frac {h^n}{n!}##, or ##1 + h + h^2/2 + h^3/6 + \cdots##. Subtracting one and dividing by h yields ##(e^h-1)/h = \sum_{n=0}^{\infty} \frac {h^n}{(n+1)!}##, or ## 1+h/2+h^2/6+\cdots##. All of those terms involving hn (n>0) vanish as h→0, leaving just 1 in the limit as h→0.
    Last edited: Sep 25, 2013
  5. Sep 25, 2013 #4


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    The problem is asking to derive an expression for the derivative of the exponential. The power series for the exponential function arises in one of two ways. Either we use the Taylor series, which depends on already knowing the derivative, or we use the definition of the exponential in terms of the limit in the same manner that I suggested. So your suggestion is valid (barring the typos in the expression ## e^h = 1 + h + h/2 + h^2/6 + \cdots##), but the pedagogy is a bit mixed up. I gave the method most appropriate to a typical course presentation, where the limit definition of the exponential comes before the discussion of the derivative, with infinite series for arbitrary functions being discussed after differential calculus.
  6. Sep 25, 2013 #5


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    Honestly then someone asks why
    [tex] e^h = \lim_{n\to \infty} (1+h/n)^n [/tex]

    and there's just way too much math involved. I prefer the alternative (below is not 100% rigorous thing, but I think it's clear that it works): Suppose there exists some number e such that
    [tex] \lim_{h\to 0} \frac{ e^h - 1}{h} = 1 [/tex].
    In this case, when h is very small
    [tex] e^h - 1 \approx h [/tex]
    [tex] e \approx (1+h)^{1/h} [/tex]

    So e is going to be
    [tex] \lim_{h\to 0} (1+h)^{1/h} [/tex]
    which is easily seen to be equivalent to the usual limit definition of e. So instead of taking this mysterious number and being amazed that it works (and being unsure of why), we start by trying to find the right base to work with and finding out that it's the Euler number.

    Also, as the thread has very little to do with algebra, I'm going to move it to the calculus forum.
    Last edited: Sep 25, 2013
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