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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

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    I don't think your "answer" is really an answer. It might make sense added to the *question itself* rather than as a separate answer, or as a comment. But you aren't answering the question posed, though.2011-04-27

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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.

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    But if this is actually something that your *learned* from one of the answers given already, then Arturo is right, you should probably post it as a comment on the answer that showed you the "way".2011-04-27
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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 $\lim_{x\to\pm\infty}\left(1+\frac{a}{x}\right)^x=\lim_{x\to\pm\infty}\left[\left(1+\frac{1}{\frac{x}{a}}\right)^\frac{x}{a}\right]^{\frac{a}{x}\cdot x}=e^a.$

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    +1, This is a great explanation for teaching students!! I'll remember it for next time I TA undergraduate calculus.2011-04-27
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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 . $

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    I have a humorous name for this: explogging. Got an intransigent limit? Explog!2011-04-29
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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.