prove (lim n -> infinity) (1 + 1/n)^n = e


by kishtik
Tags: 1 or nn, infinity, proof, prove
kishtik
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#1
Jan13-05, 08:47 AM
P: 115
Would you please proove that

(lim n -> infinity) (1 + 1/n)^n = e ?
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Galileo
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#2
Jan13-05, 09:03 AM
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For me, that is the definition of the number e.

So if you want to prove it, you have to tell what your definition of e is to show it is equivalent to that limit.

If for example:

[tex]\mbox{ e is the number such that:} \quad \lim_{h \to 0} \frac{e^h-1}{h}=1[/tex]
is your definition. Then take the derivative of ln(x) at the point 1. (Expressed as a limit)
HallsofIvy
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#3
Jan13-05, 09:09 AM
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PF Gold
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Using what preliminaries? That is sometimes used as the definition of e!

Lacking that, the simplest way is to use L'Hopital's rule. Strictly speaking L'Hopital's rule only applies to limits of differentiable functions but the limit of (1+1/x)<sup>x</sup>, as x-> infinity, must be the same as the limit of the sequence.

At x= infinity this is of the form "1<sup>infinity</sup>" so let y= (1+ 1/x)<sup>x</sup> and take the logarithm: ln(y)= x ln(1+ 1/x) is now of the form "infinity*0". We can write that as \[\frac{ln(1+1/x)}{x}\] and differentiate numerator and denominator separately. The derivative of ln(1+1/x) is \[\frac{1}{1+1/x}(-\frac{1}{x^2})= -\frac{1}{x^2+x}\]. The derivative of 1/x is -1/x<sup>2</sup>. The ratio of those two is \[\frac{x^2}{x^2+x}= \frac{x}{x+1}= \frac{1}{1+\frac{1}{x}}\] and the limit of that, as x-> infinity, is 1.

Since the limit of ln y(x)= 1, and ln is continuous, the limit of y itself is e.

kishtik
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#4
Jan13-05, 09:50 AM
P: 115

prove (lim n -> infinity) (1 + 1/n)^n = e


Thanks but I couldn't understand the latex code.
vincentchan
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#5
Jan13-05, 04:39 PM
P: 611
Okay, I'll use the definition in calculas to PROVED that.....

[tex] \frac{d}{dx} e^x = e^x [/tex] -------->>>difinition of e
substitude [tex] e = \lim (1+1/n)^n [/tex] into [tex] e^x[/tex]

we have [tex] e^x = \lim (1 + 1/n)^{nx} [/tex]..

let [tex]u=nx[/tex],

we got [tex]e^x = \lim (1 + 1/n)^{nx}=\lim (1 + x/u)^u=\lim (1 + x/n)^n[/tex]

[tex] \frac{d}{dx}\lim (1 + 1/n)^{nx} = \frac{d}{dx}\lim (1 + x/n)^n = \lim n(1+x/n)^{n-1}*(1/n) = \lim (1 + x/n)^n=\lim (1 + 1/n)^{nx}[/tex]

so..[tex] e=\lim (1 + 1/n)^n [/tex]
HallsofIvy
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#6
Jan13-05, 07:55 PM
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PF Gold
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Oh blast! I forgot which forum I was in!

At x= infinity this is of the form "1infinity" so let y= (1+ 1/x)x and take the logarithm: ln(y)= x ln(1+ 1/x) is now of the form "infinity*0". We can write that as [itex]\frac{ln(1+1/x)}{x}[/itex] and differentiate numerator and denominator separately. The derivative of ln(1+1/x) is [itex]\frac{1}{1+1/x}(-\frac{1}{x^2})= -\frac{1}{x^2+x}[/itex]. The derivative of 1/x is -1/x2. The ratio of those two is [itex]\frac{x^2}{x^2+x}= \frac{x}{x+1}= \frac{1}{1+\frac{1}{x}}[/itex] and the limit of that, as x-> infinity, is 1.

Since the limit of ln y(x)= 1, and ln is continuous, the limit is e.


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