5

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

Community
  • 1
Paul Manta
  • 3,505
  • What you have is incorrect... The limit of the above is $\infty$. Perhaps you meant $\displaystyle \lim_{x \rightarrow \infty} \left( 1 + \frac{a}{x} \right)^x$ in which case it is (one of) the definitions of $e^a$. –  Apr 27 '11 at 18:31
  • Sorry. It was a typo. I replaced the $x$ with $1$. – Paul Manta Apr 27 '11 at 18:32
  • You can see a proof, inter alia, in my answer here: http://math.stackexchange.com/questions/31387/please-help-me-to-show-that-ln-x-frac1-x/31396#31396 – Arturo Magidin Apr 27 '11 at 18:34
  • There was a previous question that treated $a=2$, but I can't seem to find it... – J. M. ain't a mathematician Apr 27 '11 at 18:35
  • I really don't understand how you conclude that the last line equals $e^{a+1}$. – Eric Naslund Apr 27 '11 at 18:36
  • @Paul: What is your definition of $e^a$? –  Apr 27 '11 at 18:37
  • What stops you is that the argument in the first limit is doing a change of variable $u = \frac{x}{a}$ to rewrite the inner limit as $(1 + \frac{1}{u})^u$, which approaches $e$. In the second limit, that change of variable does not lead to that limit because the exponent does not match the denominator in the fraction. – Arturo Magidin Apr 27 '11 at 18:37
  • @Paul: It is highly instructive to "get your hands dirty" with proofs. If you are comfortable with the identity $e = \displaystyle \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^x$, then try to mimic the proof with $\frac{1}{x}$ replaced by $\frac{a}{x}$. If you are not comfortable with that identity, then this is a different question altogether. – JavaMan Apr 27 '11 at 18:41
  • 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. – Arturo Magidin Apr 27 '11 at 19:28

4 Answers4

14

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

zar
  • 4,602
  • 2
    +1, This is a great explanation for teaching students!! I'll remember it for next time I TA undergraduate calculus. – Eric Naslund Apr 27 '11 at 22:50
7

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 $$ \mathop {\lim }\limits_{x \to \infty } \exp \bigg(x\int_1^{1 + a/x} {\frac{1}{u} \,du} \bigg) = e^a . $$

Shai Covo
  • 24,077
3

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.

Qiaochu Yuan
  • 419,620
0

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.

Arturo Magidin
  • 398,050
Paul Manta
  • 3,505
  • This should probably be a comment either on the question, or on someone's answer; not really an answer. – Arturo Magidin Apr 27 '11 at 19:27
  • I originally wanted to post it as an edit to my question. I figured it more of an answer since it explains the cause of my confusion and addresses my question (that is, why the last identity is not correct). :) Sorry if I should have posted this as a comment/edit. – Paul Manta Apr 27 '11 at 19:29
  • @Arturo, @Paul: I think this is (borderline) okay. Recognizing one's own difficulty, solving the problem, and posting an answer on one's own question is explicitly allowed in the SE framework. – Willie Wong Apr 27 '11 at 19:48
  • 2
    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". – Willie Wong Apr 27 '11 at 19:49