TITLE: Proof that this limit equals $e^a$
QUESTION [5 upvotes]: 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

REPLY [13 votes]: 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 $$\lim_{x\to\pm\infty}\left(1+\frac{a}{x}\right)^x=\lim_{x\to\pm\infty}\left[\left(1+\frac{1}{\frac{x}{a}}\right)^\frac{x}{a}\right]^{\frac{a}{x}\cdot x}=e^a.$$