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Proof lim (x+1)^(1/x)=e

  1. Aug 15, 2008 #1
    proof lim (x+1)^(1/x)=e. Where can I find the proof??
  2. jcsd
  3. Aug 15, 2008 #2


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    How are you defining "e"?
  4. Aug 15, 2008 #3
    Hmm Hi. I don't know...

    its lim (1+x)^(1/x)=e. I what to see why that is the case.

    oh e=2.718281828.....
  5. Aug 15, 2008 #4
    That's not a definition.
  6. Aug 15, 2008 #5
  7. Aug 15, 2008 #6
    You have [tex]1^\infty[/tex]. The idea is to get to 0/0 or [tex]\frac{\infty}{\infty}[/tex]. In your case, try to message your expression so that you're doing this:

    1^\infty \rightarrow 0 \cdot \infty \rightarrow \frac{0}{0}.

    You'll need to take the logarithm of both sides before you start. Once you get to 0/0, use L' Hopital's rule.
  8. Aug 16, 2008 #7


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    But the t=lnx function is defined by that it is the inverse of y=e^x, which is defined by that e=lim h --> h (1+h)^(1/h) so basically it is circular logic. The definitions must start somewhere. You can though find a different definition of e, and then connect these links.
  9. Aug 16, 2008 #8


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    If you don't know the definition of e, you can't possibly prove something is equal to it!

    there are, in fact, many different ways to define e and how you would prove something is equal to e depends strongly on your definition. For example, that limit can, very reasonable, be given as the definition of e, just as Bright Wang (and you) said. In that case, there is nothing to prove.

    My preference is to first define
    [itex]ln(x)= \int_0^x\frac{1}{t}dt[/itex]

    One can then prove that ln(x) is a one-to-one function, from all non-negative real number to all real numbers and so has an inverse. If you define exp(x) to be that inverse, it is not to difficult to prove that exp(x)= (exp(1))x. You can then define e to be exp(1)- that is, that ln(e)= 1.

    Now, you can "take logarithms of both sides and use L'Hopital's rule" as triangleman said.
    Last edited by a moderator: Aug 16, 2008
  10. Aug 19, 2008 #9
    Ill prove it for ya!

    Lets say that we have some variable called "y" and lets set it equal to (1+x)^(1/x).
    so far we have:


    now lets take the natural log of each side to obtain


    now, we can pull out the "1/x" term in front of the "Ln(" [because of your basic log rules] to get:

    Ln(y) = (1/x)*Ln(1+x) ~ or ~ Ln(y)=Ln(1+x)/x

    now we get to take the limit (yay). this leaves

    Ln(y) = lim Ln(1+x)/x
    ~~~~~ x->0

    now we get to use La Hôpital's Rule on the right hand side of the equation.

    = lim (d/dx[Ln(1+x)]) / (d/dx[x])

    (where d/dx[f(x)] refers the the derivative of f(x) with respect to "x")

    = lim (d/dx[1+x]/(1+x)) / 1
    = lim 1/(1+x)

    This last Limit can be simply evaluated by just plugging in 0 for "x"

    = lim 1/(1+0) = 1

    So! remembering the left hand side of the equation, we have

    Ln(y) = lim Ln(1+x)/x

    Ln(y)=1 ~and~ y = e^1 = e
    Thus: Lim (x+1)^(1/x)=e
    Last edited: Aug 19, 2008
  11. Aug 19, 2008 #10
    Someone is going to have my head for this, but here goes anyway:
    (We already know)
    One thing that is so special about e is the fact that if we use it to exponentiate a real variable ([tex] e^x [/tex] say) and we take the derivative of it, we end up with the same function.
    (/We already know)

    So lets define e as the number that has this property, we don't yet know what it is but here goes:

    [tex] \frac{d e^x}{dx} = e^x [/tex]
    From first principles:

    [tex] \frac{d e^x}{dx} = \lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h} [/tex]

    Now from what we desire:

    [tex] \frac{d e^x}{dx} = e^x [/tex]


    [tex] e^x = \lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h} [/tex]

    [tex] e^x = \lim_{h\rightarrow 0}\frac{e^{x}e^{h}-e^x}{h} [/tex]

    Factoring out the common [tex]e^x[/tex]

    [tex] e^x = \lim_{h\rightarrow 0}e^x\left( \frac{e^{h}-1}{h} \right)[/tex]

    [tex] e^x = e^x \lim_{h\rightarrow 0}\frac{e^{h}-1}{h} [/tex]

    This is true iff:

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

    [tex] \lim_{h\rightarrow 0}e^{h}-1 = \lim_{h\rightarrow 0}h [/tex]

    [tex] \lim_{h\rightarrow 0}e^{h} = \lim_{h\rightarrow 0}h+1 [/tex]

    [tex] e = \lim_{h\rightarrow 0}(h+1)^{\frac{1}{h}} [/tex]

    I abused the limiting process at the end there, and took the '1/h root' but it achieves the desired result. :smile:

    This by no means a rigorous proof as I'm not even sure I'm 'allowed' to perform the operations that I did at the end with the limits. My epsilon-delta skills are not that honed I'm afraid, so this is the best I could do.
    Last edited: Aug 19, 2008
  12. Aug 19, 2008 #11


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    That's a perfectly good proof Eidos- provided you have already proved that
    [tex]\frac{de^x}{dx}= e^x[/tex]
    without using that limit. And you can do that if you start from the right definition of e.
  13. Aug 19, 2008 #12
    Could you use this as the definition for e?
    [tex]\frac{de^x}{dx}= e^x[/tex]
  14. Aug 19, 2008 #13


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    Not that alone because that is a differential equation that has many solutions: y(x)= ex+ a has that property for any real number a.

    You can do the following:
    Specifying a single value removes the ambiguity.

    One possible objection to that definition is that it does not make clear that the notation ex, i.e. a number to the x power, is appropriate. However, that can then be proved. Since d ex/dx= ex, it is easy to show that the function is one-to-one, from all real numbers to all positive real numbers and so has an inverse function, from all positive real numbers to all real numbers. If we call that inverse function ln(x) then it is easy to see that d ln(x)/dx= 1/x and so
    [tex]ln(x)= \int_1^x dt/t[/tex]
    the definition I mentioned earlier.

    From that we can prove the "logarithm" property that ln(xy= y ln(x). In particular, If y= ex, then x= ln(y) so, for x non-zero, 1= (1/x)ln(y)= ln(y1/x) and, going back to the exponential form, e1= e= y1/x and then it follows that y is in fact e "to the x power".
  15. Aug 19, 2008 #14
    Ah thanks, that clears things up :smile:
  16. Aug 19, 2008 #15
    No, because then "e" could equal to Zero.
  17. Aug 21, 2008 #16
    So, can anyone prove it with an epsilon-delta proof?
  18. Aug 22, 2008 #17


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    Once again, starting from WHAT definition of "e"?
  19. Sep 23, 2008 #18
    Hey guys -

    I'm just dropping in having done a search on lim (1+1/n)^n (been too long and I've forgotten SO much)

    I answer questions on MathNerds.com and the technically "correct" answer isn't always a very "good" answer.

    In any case -

    Regarding the definition of e, Pi, e^x, ln x and the trig functions too - you might want to check out Bartle's, "The Elements of Real Analysis". It's very well written and there are "projects" that develop all of these 'taken for granted' topics mostly based on the FTC and/or basic continuity.

  20. Sep 23, 2008 #19


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    My question, which was posted 5 months ago, was really directed to the original poster. I would think it obvious that how you prove that a given limit is "= e" depends strongly on how you define e. There are several equivalent definitions and how you prove something is equal to e depends upon which definition you want to use.
  21. Jun 11, 2009 #20
    hey guys...there's another value for lim (1+x)^1/x..it goes like this...
    e(1 - x/2 + 11x^2/24 ........)
    found it in one of the books of higher math......but i cant find its proof.....can anyone help me out???
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