Proof? of the limit$ \lim_{n \to \infty}(1+\frac{1}{n})^{n} = e $ we do know the formula for e:
$$ \lim_{n \to \infty}(1+\frac{1}{n})^{n} = e $$
I was hoping to prove it using LH. Assume we don't know the value
$$\lim_{n \to \infty}(1+\frac{1}{n})^{n} = x $$
Then we can use logarithms to solve this and since $$ \ln(x) $$ is continuous everywhere we can write
$$\lim_{n \to \infty}n \ln(1+\frac{1}{n}) = \ln x $$
apply LH:
$$\lim_{n \to \infty}\frac{\ln(1+\frac{1}{n})}{\frac{1}{n}}  = \ln x $$
$$\lim_{n \to \infty}\frac{\frac{1}{(1+\frac{1}{n})}\frac{-1}{n^2}}{\frac{-1}{n^2}}  = \ln x $$
this simplifies to
$$\lim_{n \to \infty}\frac{1}{1+\frac{1}{n}} = \ln x $$
which we can evaluate to 
$$ \frac {1}{1+0} = 1 = \ln x$$
Therefore we come at
$$ \ln x = 1 $$
$$x=e$$
What I'm wondering about is whether i am actually able to use the natural logarithm because i assume i don't know what 'e' is. Is this actually validor i have to take the limit purely as a definition of the euler number?
 A: A person can define $\ln(x)$ and prove its derivative is $1/x$, etc, without knowing anything about  $(1+1/n)^n$:
For $x>0$ define $$\ln(x)=\int_1^x\frac{dt}t.$$
It follows that $\ln'(x)=1/x$. Hence $\ln$ is strictly increasing on $(0,\infty)$. And it's clear that $$\lim_{x\to\infty}\ln(x)=\int_0^1\frac{dt}t=\infty.$$Similarly $\lim_{x\to0}\ln(x)=-\infty$.
So $\ln$ is a strictly increasing differentiable bijection from $(0,
\infty)$ onto $\mathbb R$. So it has a strictly increasing differentiable inverse. Define $\exp:\mathbb R\to(0,\infty)$ to be the inverse of $\ln$. Define $e=\exp(1)$.
Note that $$\ln(xy)=\int_1^x\frac{dt}t+\int_x^{xy}\frac{dt}t=\int_1^x\frac{dt}t+\int_1^y\frac{dt}t=\ln(x)+\ln(y),$$by a simple  substitution in the last integral. Hence $\ln(x^n)=n\ln(x)$ by induction; hence $\exp(nx)=\exp(x)^n$.
Now you don't even need L'Hopital, just the definition of the derivative shows that $$\lim_{n\to\infty}n\ln(1+1/n)=\lim_{n\to\infty}\frac{\ln(1+1/n)-\ln(1)}{1/n}=\ln'(1)=1/1=1.$$ Since $\exp$ is the inverse of $\ln$ and $\exp$ is continuous this shows that $$\lim(1+1/n)^n=\lim\exp(n\ln(1+1/n))=\exp(1)=e$$.
A: All of the objects involved here ($e, \ln x$, etc.) have several equivalent definitions and it takes some interesting work to prove that they're all equivalent; until you do questions like "how do I prove X" are ambiguous because they might be easy using some definitions and hard using others (before you've proven that they're all equivalent). 
For example, you can define $\ln x$ without knowing what $e$ is, by defining it as the integral
$$\ln x = \int_1^x \frac{dt}{t}.$$
Using this integral you can prove the fundamental properties that $\ln (ab) = \ln a + \ln b$ and also that $\ln (1 + x) \approx x$ for $x$ small, which is enough to run your proof. Now you can define $e^x$ as the inverse of $\ln x$, if you want. 

My preferred way to prove all of these equivalences is to start from the definition that $e^x$ is the unique function $f(x)$ satisfying $f(0) = 1$ and $f'(x) = f(x)$. From here, you can show that
$$e^x = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$$
by showing that this limit defines a function $g(x)$ which also satisfies $g(0) = 1$ and $g'(x) = x$; hence it must be equal to $e^x$, and now you can plug in $x = 1$. This identity can be interpreted as using Euler's method with smaller and smaller step sizes to approximate the solution to $f'(x) = f(x)$. 
Generally speaking it's much nicer to prove things about $e^x$ rather than $e$, and arguably it is by far the more fundamental object. 
A: I supplied an
elementary proof
(not my own)
 that
$\lim_{n \to \infty} (1+1/n)^n$
exists here:
What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?
$e$ is then the name of that limit.
Once you know 
that the limit exists
you can then do
a variety of things
to show that it is the same
as your definition of $e$.
