
#1
Oct2209, 06:17 PM

P: 517

Is there someone who can explain why this is true, or point me to an online resource that provides a proof of it?
[tex] e^x = \lim_{n\to \infty} \left(1 + x/n \right) ^n [/tex] I know that in some ways, this is how the exponential function is defined. But any resources you can provide that explain it in more detail would be appreciated. Thanks! 



#2
Oct2209, 06:53 PM

Sci Advisor
P: 5,935

I am not sure if this will satisfy you.
However expand (1+x/n)^{n} by the binomial theorem. Take the term by term limit as n > oo. The result is the power series for e^{x}. 



#3
Oct2209, 08:16 PM

HW Helper
P: 2,149

The reason is when n is large we have
1+x/n~exp(x/n) we also have (n need not be large) exp(x/n)^n=exp(x) The proof depends on ones definition of e^x such as 1)exp'(x)=exp(x) with exp(0)=1 2)exp(x)exp(y)=exp(x+y) with exp'(0)=1 3)exp(x)=1+x+x^2/2!+...+x^n/n!+... 4)exp(log(x))=x log having been previously defined A common one given 3) is to expand (1+x/n)^n by the binomial theorem and then shown the limit of the sum is the sum of the limits. 



#4
Oct2209, 11:27 PM

P: 867

On why e^x = lim (1+x/n)^n
Let f(x) be the right side, then rewrite it as e^{lnf(x)}. With ln(f(x)), rearrange it so you can use l'Hôpital's rule (with respect to n), probably use a substitution, and then you can evaluate the limit.




#5
Oct2309, 02:47 AM

HW Helper
P: 2,149

Why does everybody like l'Hôpital's rule so much? It is not needed here as log'(1)=1 suffices.




#6
Oct2309, 07:34 AM

P: 240

I wrote the proof for you




#7
Oct2309, 12:33 PM

P: 256

As for the OP, yeah, generally, if you let the limit be a variably, say y, you could ln both sides, and solve it that way. 



#8
Nov1009, 07:18 AM

P: 509

Might be I am little late with this post...however,let me say that I could not understand how L Hospital's would have to be implemented to do this...I could follow the binomial expansion argument or the use of standard limit argument.
A second issue...From the texts that deal with symmetry,the limit is also used with operator argument.Like [tex]\displaystyle\lim_{k\to\infty}\ [\ I_{\ n\times\ n}\ + \frac{\ i \vec{\phi}\ . \mathbb{D} }{k}]^\ {k}[/tex] [tex] = \ exp\{ \ {\ i \ {\vec{\phi}\ . \mathbb{D} } \}[/tex] 



#9
Nov1009, 07:58 AM

P: 509

I am sorry that it took time to fix the Latex symbols...Let me get back to the discussion...For operator valued argument,the identity also holds...My question is will the relation still hold if [tex]\mathbb{D}=\mathbb{D}_{1}\ +\mathbb{D}_{2}[/tex] and the two operators D1 and D2 do not commute...?
I think it will(as for angular momentum matrices,which are generators of rotation),but how to see it? 



#10
Nov1009, 09:42 AM

P: 256

y = \lim_{n\to \infty} \left(1 + x/n \right) ^n [/itex] Right? So [itex] ln y = ln [ \lim_{n\to \infty} \left(1 + x/n \right) ^n] [/itex] Focus on the right, it looks kinda like this ... [itex] \lim_{n\to \infty} \left n ln [(1 + x/n \right)] [/itex] The n goes to the front because of logarithms. [itex] \lim_{n\to \infty} \left (1/n)^1 ln [(1 + x/n \right)] [/itex] If you take the limit, you get [itex]0/0[/itex] So, apply l'Hôpital's rule and you'll get [itex] ln y = \lim_{n\to \infty} \left x/(1+x/n) = x [/itex] So, cancel the ln, and you get as desired, [itex] y = e^x [/itex] 



#11
Nov1009, 11:59 AM

P: 509

But can you write in general
[itex] ln [ \lim_{n\to \infty} \left(1 + x/n \right) ^n]=\ lim_{n\to \infty} \left n ln [(1 + x/n \right)] [/itex] I vaguely remember,from outside, one can insert ln inside [lim n>0] with some condition satisfied...I cannot remember what the condition was... 



#12
Nov1009, 01:28 PM

P: 256

The condition is about continuity. ln is continuous, so yeah, you can move the limit inside or outside.




#13
Nov1009, 08:54 PM

P: 509

thanks...I forgit it...




#14
Nov1309, 08:00 PM

P: 7

Hi,
I would like to understand how l'Hôpital's rule applies. we haven't covered it yet. Thank you 



#15
Nov1309, 08:58 PM

P: 7

Hi,
We have not covered the l Hopitals rule yet so I am trying to expand (1+x/n)^n to show as lim(n>infinity) (1+x/n)^n = e^x for any x> 0 This is what I did so far and after that I am short of lost: Please read this (n 1) in vertical form ( I do not know how to use Latex) lim (n>infinity)(1 + x/n)^n = lim(x>infinity)[1 + (n 1) (1/n) + (n 2)(1/n^2) + (n 3)(1/n^3)  +   (n k)(1/n^k) + +(n n)(1/n^n)] =1 + 1/1! + 1/2! + 1/3! +  1/k! +  + 1/n! now how does this lead to e^x? May be I did not understand what you meant? Thank you. 


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