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

by glueball8
Tags: 11 or xe, proof
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glueball8
#1
Aug15-08, 01:04 PM
P: 347
proof lim (x+1)^(1/x)=e. Where can I find the proof??
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HallsofIvy
#2
Aug15-08, 01:10 PM
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How are you defining "e"?
glueball8
#3
Aug15-08, 01:12 PM
P: 347
Quote Quote by HallsofIvy View Post
How are you defining "e"?
Hmm Hi. I don't know...

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


oh e=2.718281828.....

uman
#4
Aug15-08, 04:09 PM
P: 352
Proof lim (x+1)^(1/x)=e

That's not a definition.
NoMoreExams
#5
Aug15-08, 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.
triangleman
#6
Aug15-08, 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.
disregardthat
#7
Aug16-08, 08:03 AM
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P: 1,810
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.
HallsofIvy
#8
Aug16-08, 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 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.
Math Man900
#9
Aug19-08, 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 H˘pital'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
Eidos
#10
Aug19-08, 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 epsilon-delta skills are not that honed I'm afraid, so this is the best I could do.
HallsofIvy
#11
Aug19-08, 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.
Eidos
#12
Aug19-08, 07:41 AM
P: 108
Quote Quote by HallsofIvy View Post
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.
Could you use this as the definition for e?
[tex]\frac{de^x}{dx}= e^x[/tex]
HallsofIvy
#13
Aug19-08, 09:34 AM
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PF Gold
P: 39,682
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:
f(x)= ex is defined as the function, y, satisfying the differential equation dy/dx= y, together with the initial value y(0)= 1.
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".
Eidos
#14
Aug19-08, 09:47 AM
P: 108
Ah thanks, that clears things up
Math Man900
#15
Aug19-08, 04:30 PM
P: 8
Quote Quote by Eidos View Post
Could you use this as the definition for e?
[tex]\frac{de^x}{dx}= e^x[/tex]
No, because then "e" could equal to Zero.
qspeechc
#16
Aug21-08, 04:01 PM
P: 792
So, can anyone prove it with an epsilon-delta proof?
HallsofIvy
#17
Aug22-08, 05:17 AM
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PF Gold
P: 39,682
Once again, starting from WHAT definition of "e"?
hilbert
#18
Sep23-08, 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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