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Indeterminate forms

  1. Apr 19, 2005 #1
    I have the solution to this particular problem, the problem is the person who solved it used a method that unfamiliar with. Does any one know how to solve this problem using Binomial expansion. I know that the answer is [tex] e^k[/tex] where K is some constant. this is the problem:

    [tex]\lim_{x\rightarrow\infty}(1 + \frac{k}{x})^x[/tex]
    Last edited: Apr 19, 2005
  2. jcsd
  3. Apr 19, 2005 #2


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    I'm having difficulty seeing how binomial expansion could help. (Even in its general form, which you'd have to use, because x isn't restricted to the integers)

    You can't solve the problem until you decide upon an acceptable definition of e^k, and possibly have proven some facts about it. (For example, sometimes e^k is defined to be the value of this limit! So, in that case, the proof is trivial)
  4. Apr 19, 2005 #3

    his solution starts like this:

    [tex]1 + x\frac{k}{x} + x\frac{(x-1)}{2!}+\frac{k^2}{x^2}+x\frac{(x-1)(x-2)}{3!}\frac{k^3}{x^3}[/tex]

    But i have no idea what this means.
  5. Apr 19, 2005 #4
    Here's an artifical way you could do this:

    Let the limit be L. Take natural log of both sides. This gives

    [itex]\ln L = \ln(\lim_{x\rightarrow\infty}(1 + \frac{k}{x})^x)[/tex]

    As the log of a limit is the limit of log for continuous functions, we have

    [tex]\ln L = \lim_{x\rightarrow\infty}x\ln((1 + \frac{k}{x}))[/tex]

    which is the same as

    [tex]\ln L = \lim_{x\rightarrow\infty}k\frac{\ln((1 + \frac{k}{x}))}{k/x}[/tex]

    the limit is k and so [itex]L = e^k[/itex].

    The theorem we have used here is that the log of a limit of a continuous function is the limit of the log of the function. But there is no better way than to use the definition of e, which is

    [tex]e = \lim_{n\rightarrow\infty}(1+\frac{1}{n})^n[/tex]

  6. Apr 19, 2005 #5

    Okay I get it now. "He" has used a binomial expansion for [itex](1+k/x)^x[/itex]. You should know that x can be negative, fractional or positive integral so you haave to use the binomial theorem for general nonintegral index (as done here). But there is a catch.

    The idea used here is that as the order of every term grows, the terms in the brackets partially cancel the the (1/x)^k term. In more precise terms, the at least one of the product terms reduces the power of the denominator to 0. This method looks practical but it is not mathematically correct unless you first prove that the binomial expansion written actually converges to zero for higher order terms as [itex]x\rightarrow\infty[/itex]. This is one of those things which doesn't seem to deserve special proof, but it does.

    So I would still think that the argument given by Hurkyl (and me in my first post) are closer to the acceptable definition of e.


    EDIT: By the way, the second term of your original binomial expansion is incorrect. Change the + sign to a product and it'll be okay then.

    EDIT#2: A better way to understand the order argument is as follows: a typical term of the the continued product in the numerator is of the form (x-k) for k = 0, 1, 2, ...kmax and for every such product the term in the denominator is x^kmax. Hence, there are as many terms in the denominator (I am breaking x^k as x*x*...*x (k times)). So the order of the numerator = order of denominator in terms of the powers of x. Of course, the order is same but there are still some lurking lower order terms in the numerator after cancellation. They can be reclubbed with the higher order denominator. This intuitively tells you that for large x, the sequence converges to some value. But you need to prove that it does (rigourously) first and find that value. All this is unnecessary if you are allowed to use the proof right out of the box though.
    Last edited: Apr 19, 2005
  7. Apr 19, 2005 #6
    Thank you Guys!
  8. Apr 19, 2005 #7


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    I would like to point out that method of proof has a lot of subtle technicalities. Remember the famous "proof" that [itex]\lim_{n \rightarrow \infty} n / n = 0[/itex]:

    Fake proof:
    1/1 = 1/1
    2/2 = 1/2 + 1/2
    3/3 = 1/3 + 1/3 + 1/3
    4/4 = 1/4 + 1/4 + 1/4 + 1/4
    Now, each individual term goes to zero. Therefore, the limit is simply a sum of many zeroes, and is thus zero!

    The moral is that you're interchanging the limit operations, and that's something you should only do if you really know what you're doing.

    With the binomial expansion method, you have to let the number of terms go to infinity first, and THEN let x go to infinity.

    So, the argument you posted really isn't valid unless you apply some advanced machinery that lets you do things the other way around. (Though it is a nice heuristic argument)
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