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Limit tends to infinity

  1. Apr 17, 2008 #1
    Why does
    [tex] \lim(1+\frac{1}{x})^x = e[/tex]
    [tex] x->\infty[/tex]
  2. jcsd
  3. Apr 17, 2008 #2
    This is one possible definition of the number e.

    What is your definition of e?
  4. Apr 17, 2008 #3

    O so how do i prove this ?
  5. Apr 17, 2008 #4


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

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

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

    Then use L'Hopital's Rule twice.
  6. Apr 17, 2008 #5


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    If you don't have a definition of e, then you can't.
  7. Apr 17, 2008 #6
    How do you prove what?:smile:
  8. Apr 17, 2008 #7
    yea how do you prove a definition? maybe he means how do you consistency?
  9. Apr 18, 2008 #8


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    Unfortunately, johndoe hasn't gotten back to us to answer the question about what definition of e he is using.

    Yes, many texts define e by that limit. If that is the definition, then no "proof" is required.

    Others, however, treat the derivative of f(x)= ax this way:
    [tex]f(x+ h)= a^{x+ h}= a^x a^h[/tex]
    [tex]f(x+ h)- f(x)= a^xa^h- a^x= a^x(a^h- 1)[/tex]
    [tex]\frac{f(x+h)- f(x)}{h}= a^x\frac{a^h- 1}{h}[/tex]
    Which, after you have shown that
    [tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}[/tex]
    exists, shows that the derivative of ax is just a constant times ax.
    "e" is then defined as the value of a such that that constant is 1:
    [tex]\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1[/tex]

    We can then argue (roughly, but it can be made rigorous) that if, for h close to 0, (eh-1)/h is close to 1, eh- 1 is close to h and so eh is close to h+ 1. Then, finally, e is close to (h+1)1/h. As h goes to 0, 1/h goes to infinity. letting x= 1/h, (h+1)1/h becomes (1/x+ 1)x so
    [tex]\lim_{x\rightarrow \infty}(1+ \frac{1}{x})^x= e[/tex]

    Here's a third definition of "e":
    It has become more common in Calculus texts to work "the other way". That is, define ln(x) by
    [tex]ln(x)= \int_1^x \frac{1}{t}dt[/itex]
    From that we can prove all the properties of ln(x) including the fact that it is a "one-to-one" and "onto" function from the set of positive real numbers to the set of all real numbers- and so has an inverse function. Define "exp(x)" to be the inverse function . Then we define "e" to be exp(1) (one can show that exp(x) is, in fact, e to the x power).

    Now, suppose [itex]lim_{x\rightarrow \infty} (1+ 1/x)^x[/itex] equals some number a. Taking the logarithm, x ln(1+ 1/x)must have limit ln(a). Letting y= 1/x, that means that the limit, as y goes to 0, of ln(1+y)/y must be ln(a). But that is of the form "0/0" so we can apply L'hopital's rule: the limit is the same as the limit of 1/(y+1) which is obviously 1. That is, we must have ln(a)= 1 or a= e.
    Last edited: Apr 18, 2008
  10. Apr 18, 2008 #9

    Ok I see. I didn't know that definition before until I reach a problem requiring me to find that limit.
  11. Sep 25, 2008 #10
    I believe that the definition of e is:

    e is such that when: y=e^x
    dy/dx = e^x

    Hence via the Taylor Series, the actual value of e can be found

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