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Calculus Proof

  1. Mar 1, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that the limit as h approaches 0 of ((e^h)-1) / h = ln e = 1 using two numerical examples.

    2. Relevant equations

    |Exponential

    3. The attempt at a solution

    I have the solution but don't quite understand the steps involved.
    1)(e^h-1)/h
    2)(1+h)^(1/h)=e
    3)1+h=e^h
    4).....

    I need help understanding how you get from the first step to the second step and then to the third step when proving this limit. Thanks.
     
  2. jcsd
  3. Mar 1, 2014 #2
    The steps that are in that solution don't make any sense to me, unless it is assumed that (e^h-1)/h = 1. But, by using L'Hôpital's Rule, you can use derivatives to help you find you limit if you have indeterminate forms (0/0, ∞/∞, -∞/∞, e.g.) Take a look:

    $$L = \lim_{h \to 0} \frac{e^{h}-1}{h} $$
    Since this is an indeterminate form(0/0) we can use L'Hôpital's Rule; we differentiate the numerator and denominator (separately, not as one fraction).
    $$L = \lim_{h \to 0} \frac{he^{h}}{1} $$
    $$ln(L) = \lim_{h \to 0} h ln(h) $$
    Now we need to get our limit in an indeterminate form so that we may use L'Hôpital's Rule once again..
    $$ln(L) = \lim_{h \to 0} \frac{ln(h)}{1/h} $$
    $$ln(L) = \lim_{h \to 0} \frac{1/h}{-1/h^{2}} $$
    $$ln(L) = \lim_{h \to 0} -h $$
    Now we can directly substitute 0 into our limit.
    $$ln(L) = 0$$
    $$L = e^{0} = 1$$
    Therefore, $$\lim_{h \to 0} \frac{e^{h}-1}{h} \equiv ln(e) $$
     
    Last edited: Mar 1, 2014
  4. Mar 1, 2014 #3

    Ray Vickson

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    You made some fatal errors.
    [tex] \lim_{h \to 0} \frac{e^h-1}{h} = \lim_{h \to 0} \frac{(e^h-1)'}{h'} =
    \lim_{h \to 0} \frac{e^h}{1} = e^0 = 1[/tex]
     
  5. Mar 2, 2014 #4
    2) Does not follow from 1). 2) isn't even true as written. One defintion of the number ##e## is ##e=\lim_{n\rightarrow \infty}(1+\frac{1}{n})^{n}##, which is equivalent to ##e=\lim_{h\rightarrow 0^+}(1+h)^{1/h}##. So ##(1+h)^{1/h}\approx e## when ##h## is small.

    Then if we raise each side of this non-equation to the power ##h##, we get ##1+h\approx e^h##, which is approximately what is written in 3).

    I don't know why 1) was written where it was written. I would ask the person who wrote it to clarify why they thought it was necessary to begin their proof with a non-statement that had no relevance to the statements that immediately followed. And then I would ask them why they are telling you lies like ##(1+h)^{1/h}=e##.

    I'd say the most egregious error is using ##(e^h)'=e^h## to prove that ##\lim_{h\rightarrow 0}\frac{e^h-1}{h}=1## when it is almost certain that ##\lim_{h\rightarrow 0}\frac{e^h-1}{h}=1## is, in short order, going to be used to show that ##(e^x)'=e^x##. Circular logic is bad, m'kay.
     
  6. Mar 2, 2014 #5

    Ray Vickson

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    I agree, but that was not the point. I was correcting a serious error in a previous poster's use of l'Hospital's rule.

    However, everything depends on what definition of ##e^x## is allowed, and we are not told anything about that. Perhaps the most straightforward definition is as the inverse of the ln function, but some writers have, instead, defined it as the usual infinite series. The most intricate and difficult definition seems to be to first define ##e## as the usual limit, and to then try to define ##e^x## as an 'exponentiation' operation. It can be done but it is far from trivial. It may be intuitive, but intuition is not proof.
     
  7. Mar 2, 2014 #6
    Yes. I was trying to add to your comment rather than comment on your comment. I suppose that wasn't clear.

    I agree that it totally boils down to how one defines ##e##. The textbook that my department currently uses (Briggs/Cochran) takes the easy way out and defines ##e## to be the unique number satisfying ##\lim_{h\rightarrow 0}\frac{e^h-1}{h}=1## (which is totally lazy and unfair to the students if you ask me). Regardless of how you define ##e## or ##e^x##, there is always something to prove, and it's usually non-trivial.
     
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