Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

share|cite|improve this question
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 $e^a$. – user17762 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:… – 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. 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

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

share|cite|improve this answer
+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

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

share|cite|improve this answer
I have a humorous name for this: explogging. Got an intransigent limit? Explog! – ncmathsadist Apr 29 '11 at 0:39

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.

share|cite|improve this answer
up vote 0 down vote accepted

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.

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.