Expanding the Exponential Function Using Limits

[AxiomOfChoice]

Is there someone who can explain why this is true, or point me to an online resource that provides a proof of it?
$$e^x = \lim_{n\to \infty} \left(1 + x/n \right) ^n$$
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!

 

[mathman]

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.

 

[lurflurf]

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.

 

[Bohrok]

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.

 

[lurflurf]

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

 

[elibj123]

I wrote the proof for you

 

[l'Hôpital]

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

Are you kidding? l'Hôpital's rule is the best. Made of epic win.

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.

 

[neelakash]

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

$$\displaystyle\lim_{k\to\infty}\ [\ I_{\ n\times\ n}\ + \frac{\ i \vec{\phi}\ . \mathbb{D} }{k}]^\ {k}$$

$$= \ exp\{ \ {\ i \ {\vec{\phi}\ . \mathbb{D} } \}$$

 

[neelakash]

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 $$\mathbb{D}=\mathbb{D}_{1}\ +\mathbb{D}_{2}$$ 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?

 

[l'Hôpital]

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.

$y = \lim_{n\to \infty} \left(1 + x/n \right) ^n$

Right? So
$ln y = ln [ \lim_{n\to \infty} \left(1 + x/n \right) ^n]$

Focus on the right, it looks kinda like this ...

$\lim_{n\to \infty} \left n ln [(1 + x/n \right)]$

The n goes to the front because of logarithms.

$\lim_{n\to \infty} \left (1/n)^-1 ln [(1 + x/n \right)]$

If you take the limit, you get $0/0$ So, apply l'Hôpital's rule and you'll get

$ln y = \lim_{n\to \infty} \left x/(1+x/n) = x$

So, cancel the ln, and you get as desired,

$y = e^x$

 

[neelakash]

But can you write in general

$ln [ \lim_{n\to \infty} \left(1 + x/n \right) ^n]=\ lim_{n\to \infty} \left n ln [(1 + x/n \right)]$

I vaguely remember,from outside, one can insert ln inside [lim n-->0] with some condition satisfied...I cannot remember what the condition was...

 

[l'Hôpital]

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

 

[neelakash]

thanks...I forgit it...

 

[mrm0607]

Hi,

I would like to understand

how l'Hôpital's rule applies. we haven't covered it yet.

Thank you

 

[mrm0607]

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.