How did Euler and Bernoulli prove this limit? How did the Euler and Bernoulli know that $\lim_{x \to \infty} (1+\frac1x)^x$ exists?
 A: Let $A$ be a constant such that the derivative of $f(t) = A^t$ at $t = 0$ is less than one.  Then for $t$ sufficiently small we have $1 + t > A^t$.  Likewise if the derivative of $B^t$ is greater than one at $t=0$ then for $t$ small we have $1+t < B^t$.  Setting $t = \frac{1}{x}$ shows that $A < (1+\frac{1}{x})^x < B$ for $x$ sufficiently large.  So the limit is equal to the unique constant $e$ such that the derivative of $e^t$ at $t = 0$ is actually equal to one (my favorite definition of $e$).
I don't know how Euler proved it but if you want to find out you should read his book "Introduction to Analysis of the Infinite".
A: Your edited question is simply, how do we know that 
$$\lim_{x\rightarrow\infty} (1+1/x)^x$$ exits?  In modern terminology, we might define $f(x)=(1+1/x)^x$, show that $f$ is increasing and bounded above, and deduce that the limit is the supremum of the set of function values for positive $x$, i.e.
$$\lim_{x\rightarrow\infty} f(x) = \sup \{f(x):x>0\}.$$
But, honestly, I think that Euler and his contemporaries largely took this kind of thing for granted, feeling that no proof was necessary.
A: Expand both sides in a Taylor series to prove that they are equal. 
This is the same idea as used to show $e^{ix} = \cos(x)+i\sin(x)$. If sine is expanded as a Taylor series, multiply it by $i$, and then added to the Taylor series for cosine, it equal the Taylor series for $e^{ix}$.
A: It would be tedious before computers, but you can find that limit simply by plugging in some large numbers. If n = 100 you get 2.704. If n = 1000 you get 2.716. If n = 10000 you get 2.718. Plug in 69! (Wolfram Alpha won't go much higher) and you have 20 or so decimal places that won't change. The digits that already exist won't change no matter how high n gets. It follows that if n is infinite, then the resulting number has reached a point where the decimal places aren't changing at all.
The result takes the form of $\frac{(x+1)^x}{x^x}$, which may be somewhat easier to expand. I can offer something of a heuristic proof for a limit with an example in this — notice how as the numerator gets larger, the exponent of the denominator gets significantly larger with it.
