Proof that this limit equals $e^a$ Can someone please explain to me why the following identity is true?
$$\lim_{x \to \infty}\left(1 + \frac{a}{x} \right)^x = e^a$$
(I'll make a notation $L$ that is equal to the limit above.)
One 'proof' I saw went something like this:
$$L = \lim_{x \to \infty}\left(\left(1 + \frac{a}{x} \right)^\frac{x}{a}\right)^a = e^a$$
That can't be right... right? Because there really is nothing stopping me from saying
$$L = \lim_{x \to \infty}\left(\left(1 + \frac{a}{x} \right)^\frac{x}{a + 1}\right)^{a + 1} = e^{a + 1}$$
but that's obviously not true.

Edit: I posted my own answer to this question, where I explain what got me confused:
► http://math.stackexchange.com...35491#35491
 A: Perhaps you'll find instructive the following approach.
For $a>0$,
$$
\bigg(1 + \frac{a}{x}\bigg)^x  = \exp \bigg(x\int_1^{1 + a/x} {\frac{1}{u} \,du} \bigg).
$$
Since
$$
\frac{a}{{x + a}} = \int_1^{1 + a/x} {\frac{1}{{1 + a/x}}\,du}  \le \int_1^{1 + a/x} {\frac{1}{u}\,du}  \le \int_1^{1 + a/x} {\frac{1}{1}\,du}  = \frac{a}{x},
$$
we have
$$
\frac{{xa}}{{x + a}} \le x\int_1^{1 + a/x} {\frac{1}{u}\,du}  \le a.
$$
Thus, the expression in the middle tends to $a$ as $x \to \infty$, leading to 
$$
\mathop {\lim }\limits_{x \to \infty } \exp \bigg(x\int_1^{1 + a/x} {\frac{1}{u} \,du} \bigg) = e^a .
$$
A: The proof you saw is correct. I don't understand your last equation, since it is false that $\lim_{x \to \infty} \left( 1 + \frac{a}{x} \right)^{ \frac{x}{a+1} } = e$. You need to make the substitution $y = \frac{x}{a}$ and then hopefully everything will be clear. 
A: You have to recall the fundamental limit $$\lim_{x\to\pm\infty}\left(1+\frac{1}{x}\right)^x=e.$$
Think of it as a general rule like this:
$$\lim_{\star\to\pm\infty}\left(1+\frac{1}{\star}\right)^\star=e,$$ where the star can be substituded by any expression (which tends to $\pm\infty$).
So $$\lim_{x\to\pm\infty}\left(1+\frac{a}{x}\right)^x=\lim_{x\to\pm\infty}\left[\left(1+\frac{1}{\frac{x}{a}}\right)^\frac{x}{a}\right]^{\frac{a}{x}\cdot x}=e^a.$$ 
A: I was under the impression that $\lim_{x \to \infty}\left(1 + \frac{a}{x}\right)^{x/y} = e$, regardless of what constant $y$ is. My confusion came from the fact that usually $\lim_{x \to \infty} x/y = \infty$, regardless of $y$. I now know that this is a special case and it is specifically required that $y = a$ and the power be $x/a$ and nothing else.
