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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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  • $\begingroup$ 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 $e^a$. $\endgroup$ – user17762 Apr 27 '11 at 18:31
  • $\begingroup$ Sorry. It was a typo. I replaced the $x$ with $1$. $\endgroup$ – Paul Manta Apr 27 '11 at 18:32
  • $\begingroup$ You can see a proof, inter alia, in my answer here: math.stackexchange.com/questions/31387/… $\endgroup$ – Arturo Magidin Apr 27 '11 at 18:34
  • $\begingroup$ There was a previous question that treated $a=2$, but I can't seem to find it... $\endgroup$ – J. M. is a poor mathematician Apr 27 '11 at 18:35
  • $\begingroup$ I really don't understand how you conclude that the last line equals $e^{a+1}$. $\endgroup$ – Eric Naslund Apr 27 '11 at 18:36
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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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    $\begingroup$ +1, This is a great explanation for teaching students!! I'll remember it for next time I TA undergraduate calculus. $\endgroup$ – Eric Naslund Apr 27 '11 at 22:50
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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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    $\begingroup$ I have a humorous name for this: explogging. Got an intransigent limit? Explog! $\endgroup$ – ncmathsadist Apr 29 '11 at 0:39
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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.

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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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  • $\begingroup$ This should probably be a comment either on the question, or on someone's answer; not really an answer. $\endgroup$ – Arturo Magidin Apr 27 '11 at 19:27
  • $\begingroup$ 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. $\endgroup$ – Paul Manta Apr 27 '11 at 19:29
  • $\begingroup$ @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. $\endgroup$ – Willie Wong Apr 27 '11 at 19:48
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    $\begingroup$ 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". $\endgroup$ – Willie Wong Apr 27 '11 at 19:49

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