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

Limit problem.

  1. Nov 23, 2009 #1
    Noncircular proof that the limit:

    [tex]\lim_{h \to 0}\frac{e^{h}-1}{h}[/tex]

    Tends to 1, which is [tex](ln(e))[/tex] as [tex]{h \to 0}[/tex]

    Been trying logarithms and other kind of stuff but always seem to get 0/0 :(
     
    Last edited: Nov 24, 2009
  2. jcsd
  3. Nov 24, 2009 #2
    L'Hospital gives the desired result.
     
  4. Nov 24, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, but he asked for a "non-circular" proof! The point is that the standard proof of the derivative of ex requires taking that limit. Thus, using the derivative to get that limit is "circular".

    However, it is not necessary to use the "standard proof" of the derivative. One thing commonly done in modern Calculus books is to define the logarithm first:
    Define
    [tex]ln(x)= \int_1^x \frac{1}{t} dt[/tex]

    All of the usual properties of the natural logarithm can be derived from that including the fact that it is a one-to-one function from the postive real numbers to the real numbers and so has an inverse function from the real numbers to the positive real numbers. Call that inverse function "exp(x)". Since it is clear from the definition above that the derivative of ln(x) is 1/x, knowing that ln(exp(x))= x tells us that (1/exp(x))(exp(x))'= 1 so the derivative of exp(x) is exp(x) itself. Once we have that, then we can use the L'Hospital as Matthollyw00d says.

    (Also, if y= exp(x), then x= ln(y) and, if [itex]x\ne 0[/itex] 1= (1/x)ln(y)= ln(y1/x. Going back to the exp form, exp(1)= y1/x so y= (exp(1))x. That is, the "exp" function, defined as the inverse to ln(x), really is ex where e is defined as exp(1), the number whose natural logarithm is 1.)

    But if you want a more direct proof, you have to go back to the definition of "e". e can be defined by [itex]e= \lim_{h\to 0}(1+ h)^{1/h}[/itex]. So we can say that, for h close to 0, we have [itex]e= (1+h)^{1/h}[/itex], approximately. Then [itex]e^h= 1+h[/itex] so [itex]e^h-1 = h[/itex] and then [itex](e^{h}-1)/h= 1[/itex]. That is, as I said, approximate. taking the limit, as h goes to 0, makes it exact.
     
  5. Nov 24, 2009 #4
    I love math. Im a pilot but can still sit and study math for several hours. Problem is that I don't have anyone to share my thoughts with and discuss. However you guys seem to be really bright in math. Best math/physics forum so far!

    Thank you for the reply guys!
     
  6. Nov 25, 2009 #5
    I seem to suggest using series all the time but the Taylor series expansion of e^h might be useful.
    Then (e^h-1)/h = 1+h(some terms) tends to 1 as h goes to 0. But this may again be circular.
     
  7. Nov 25, 2009 #6

    lurflurf

    User Avatar
    Homework Helper

    This limit is the definition (Newton quotient) of exp'(0).
    Thus L'Hospital is overkill.
    What we need is a definition of exp where either the limit or exp'(0) is given or easy to deduce.
    In fact exp'(0)=1 is what is special about e.
     
  8. Nov 25, 2009 #7
    limit means not simply substituting the values but it merely means that at its neighbourhood
     
  9. Nov 26, 2009 #8
    Ok something is wrong here.

    [tex]\frac{d}{dx}{a^x}=ln({a}){a^x}[/tex]

    Tis is by proving that the limit of (ah-1)/h tends to ln(a).

    The limit above with e instead: (eh-1)/h tends to 1. But it should actually tend to ln(e) wich happens to be 1. This according to "(ah-1)/h tends to ln(a)" statement which is true.

    Does this mean that L'Hospitals rule was just a coincidence and doesent actually work with (ah-1)/h since ln(a)[tex]\neq[/tex] 1 if ofcourse a[tex]\neq[/tex]e

    The derivative of ex is actually ln(e)ex, same applies for ax, then dy/dx(ax) = ln(a)ax.

    Its kind of a contradiction there if you use L'hospital to prove that these limits tend to ln(e) respectively ln(a).

    Does anyone understand what I mean or is my reasoning idiotic?
     
  10. Nov 26, 2009 #9
    Actually as someone suggested, it is very easy using the definition of e^x as the infinite power series:

    [tex] \displaystyle\sum_{k=0}^\infty \frac{x^k}{k!} [/tex] . Try taking the derivative and you will see it is the same sum, so that D(e^x) = e^x. Then you may use L'hopital's rule, whose derivation does not depend in any way on e^x (uses generalized mean value theorem), so no circular argument is used.
     
  11. Nov 26, 2009 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I don't understand what "contradiction" you are talking about. IF you have proven the derivative of ex without using that limit itself, then, yes, L'Hospital's method works nicely for either ex or ax.

    Assume that we already know that d(ex)/dx= ex and that d(ax/dx= ln(a)ax. Then Applying L'Hospital to [itex]lim_{h\to 0}(e^h- 1)/h[/itex], differentiating both numerator and denominator, with respect to h, separately, we get [itex]\lim_{h\to 0}e^h/1= 1[/itex]. Doing the same with [itex]\lim_{h\to 0}(a^h-1)/h[/itex] we get [itex]\lim_{h\to 0}ln(a)a^h/1= ln(h)[/itex], both of which are correct.
     
  12. Dec 8, 2009 #11
    What does exactly 'non-circular' proof mean?
     
  13. Dec 8, 2009 #12
    Means that you shouldnt use things you know in forehand to proof something. Sounds weird but here is an example:

    Lets say you want to proove that the derivative of the function f(x)=ax is f'(x)ln(a)ax by using the definition of the derivative

    f'(x)=(a(x+h)-ax)/h as h tends to 0.

    After simplifying you will get ax times the limit as h tends to 0 of (ah-1)/h.

    You can prove that this limit tends to ln(a) by using L'Hopitals rule, but then you have to differentiate ah, however, how can you differentiate it, if you havent prooven its derivative yet?

    see my point?
     
  14. Dec 9, 2009 #13
    jjjjj
     
  15. Dec 9, 2009 #14

    Yeah...I got it...Best Regards...
     
  16. Dec 9, 2009 #15
    did we get an answer? :confused: is [itex]\lim_{x \rightarrow 0} \frac{ln(1+x)}{x} = 1[/itex] allowed or not? because if that's allowed write [itex]a^x = e^{xlna}[/itex] & make the substitution [itex]e^{xlna} - 1 = y[/itex] so that [itex]x = \frac{lna}{ln(y+1)}[/itex]. Then [tex]\lim_{x \rightarrow 0} \frac{a^x - 1}{x} = \lim_{y \rightarrow 0} \frac{ylna}{ln(y+1)} = lna[/tex]
     
    Last edited: Dec 9, 2009
  17. Dec 9, 2009 #16
    Yes it is.

    L'Hopitals rule gives you:


    [tex]\lim_{x \rightarrow 0}\frac{1/(1+x)}{1} = 1[/tex]
     
  18. Dec 10, 2009 #17

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, the point was that whether or not that is "allowed" depends upon exactly how you have defined the logarithm. If you used the exponential to define the logarithm (ln(x) is the inverse function to ex), and derived the derivative of the logarithm from the derivative of the exponential, then you cannot use properties of the logarithm to find that limit. Of course, as I said before, you don't have to use the exponential to define the logarithm. Many texts do it the other way: define the logarithm function to be
    [tex]ln(x)= \int_1^x \frac{1}{t} dt[/tex]

    and then define the exponential to be the inverse function to the logarithm. That way you do not use [itex]\lim_{x\to 0}(e^x- 1)/x[/itex] to define the derivative of the exponential (and then the derivative of the logarithm) so there is no "circular reasoning" involved in using the logarithm to prove that limit.
     
  19. Dec 10, 2009 #18
    To differentiate term by term you first need to prove the series converges uniformly (I think it's easier in the case of power series).

    Another way to use this power series (and btw the series itself derives from the original definition of e) is to just substitute it into the limit, and since you know it converges you can work with it with no worries. You will end up with a series which's first term is 1 and all of the rest are powers of h, so when h tends to zero the series tends to 1.
     
  20. Dec 10, 2009 #19
    is that what was meant by no circular reasoning? I thought it just meant don't use l'Hopital's rule
     
  21. Dec 11, 2009 #20
    When I took Calculus we learned the power rule for obtaining the derivative. d(X^n)/dX = nX^(n-1). This carries over to negative exponents. So that now we knew, in general how to integrate X^-a, for integers and then fractions.

    BUT the whole matter breaks down in attempting to integrate 1/X. Thus my class proceeded in the way written above by HallsofIvy defining the log of X.

    is that what was meant by no circular reasoning? I thought it just meant don't use L'Hopital's rule


    We could proceed to use the Taylor expansion e^X = 1+X/1 +X^2/2! if we were to get this by expanding the difinition of e^x = lim(1+x/n)^n as n goes to infinity.

    L'Hospital's rule employs the derivative, so that if we assume that derivative and at no previous point prove it, then it is circular.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limit problem.
  1. Limit problem (Replies: 6)

  2. A limit problem (Replies: 9)

  3. Limit problem (Replies: 2)

  4. Limit Function Problem (Replies: 5)

  5. Problem with limit (Replies: 4)

Loading...