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Proof lim (x+1)^(1/x)=e 
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#1
Aug1508, 01:04 PM

P: 347

proof lim (x+1)^(1/x)=e. Where can I find the proof??



#2
Aug1508, 01:10 PM

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PF Gold
P: 39,348

How are you defining "e"?



#3
Aug1508, 01:12 PM

P: 347

its lim (1+x)^(1/x)=e. I what to see why that is the case. x>0 oh e=2.718281828..... 


#4
Aug1508, 04:09 PM

P: 352

Proof lim (x+1)^(1/x)=e
That's not a definition.



#5
Aug1508, 05:41 PM

P: 626

You should read http://en.wikipedia.org/wiki/E_(mathematical_constant) but try taking ln of both sides and using l'hopitals.



#6
Aug1508, 06:58 PM

P: 12

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:
[tex] 1^\infty \rightarrow 0 \cdot \infty \rightarrow \frac{0}{0}. [/tex] You'll need to take the logarithm of both sides before you start. Once you get to 0/0, use L' Hopital's rule. 


#7
Aug1608, 08:03 AM

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P: 1,742

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.



#8
Aug1608, 09:11 AM

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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 onetoone function, from all nonnegative 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. 


#9
Aug1908, 12:43 AM

P: 8

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: y=(1+x)^(1/x) now lets take the natural log of each side to obtain Ln(y)=Ln((1+x)^(1/x)) 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]) x>0 (where d/dx[f(x)] refers the the derivative of f(x) with respect to "x") = lim (d/dx[1+x]/(1+x)) / 1 x>0 = lim 1/(1+x) x>0 This last Limit can be simply evaluated by just plugging in 0 for "x" = lim 1/(1+0) = 1 x>0 So! remembering the left hand side of the equation, we have Ln(y) = lim Ln(1+x)/x ~~~~~x>0 Ln(y)=1 ~and~ y = e^1 = e Thus: Lim (x+1)^(1/x)=e Q.E.D 


#10
Aug1908, 02:46 AM

P: 108

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] Therefore [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. 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 epsilondelta skills are not that honed I'm afraid, so this is the best I could do. 


#11
Aug1908, 05:43 AM

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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. 


#12
Aug1908, 07:41 AM

P: 108

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


#13
Aug1908, 09:34 AM

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Not that alone because that is a differential equation that has many solutions: y(x)= e^{x+ a} has that property for any real number a.
You can do the following: One possible objection to that definition is that it does not make clear that the notation e^{x}, i.e. a number to the x power, is appropriate. However, that can then be proved. Since d e^{x}/dx= e^{x}, it is easy to show that the function is onetoone, 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(x^{y}= y ln(x). In particular, If y= e^{x}, then x= ln(y) so, for x nonzero, 1= (1/x)ln(y)= ln(y^{1/x}) and, going back to the exponential form, e^{1}= e= y^{1/x} and then it follows that y is in fact e "to the x power". 


#14
Aug1908, 09:47 AM

P: 108

Ah thanks, that clears things up



#15
Aug1908, 04:30 PM

P: 8




#16
Aug2108, 04:01 PM

P: 792

So, can anyone prove it with an epsilondelta proof?



#17
Aug2208, 05:17 AM

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PF Gold
P: 39,348

Once again, starting from WHAT definition of "e"?



#18
Sep2308, 02:14 AM

P: 1

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. Larry 


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