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I am using ${\displaystyle \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^{n}}=e^x$ and want to show that $e^{x+y} =e^xe^y$ and $e^{-x}=(e^x)^{-1}$ using only the given limit.

The trouble is that I believe $e^{x+y}=e^xe^y$ is not true for large values of $n$

grayQuant
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    That limit equation is always true. In particular it's true when you replace the $x$'s with $z$'s: where you define $z=x+y$. Or when you define $z=-1$. –  Mar 19 '15 at 03:35
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    The equation $e^{x+y} = e^{x}e^{y}$ does not depend on an integer $n$. Don't confuse the limit with the terms in the limit. – Glen M Wilson Mar 19 '15 at 03:35

2 Answers2

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Since $$\left(1+\frac{x}{n}\right)^n\cdot\left(1+\frac{y}{n}\right)^n = \left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n=\left(1+\frac{x+y}{n}\right)^n\cdot\left(1+\frac{xy}{n^2+nx+ny}\right)^n$$ we just need to show that for any $x,y$ we have: $$\lim_{n\to +\infty}\left(1+\frac{xy}{n^2+nx+ny}\right)^n=1\tag{1}$$ that follows from: $$\lim_{n\to +\infty}\left(1+\frac{xy}{n^2}\right)^n=1.\tag{2}$$

Jack D'Aurizio
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Case $1$: Assume $x+y>0$

Write $e^{x+y}={\displaystyle \lim_{n \to \infty} \left( 1 + \frac{(x+y)}{n} \right)^{n}}$ and substitute $\frac1m=\frac{x+y}{n}$.

$$e^{x+y}={\displaystyle \lim_{n \to \infty} \left( 1 + \frac{(x+y)}{n} \right)^{n}}={\displaystyle \lim_{m \to \infty} \left( 1 + \frac{1}{m} \right)^{m(x+y)}}=\left({\displaystyle \lim_{m \to \infty} \left( 1 + \frac{1}{m} \right)^{m}}\right)^{x+y}=e^xe^y$$

which completes the proof for Case 1, $x+y>0$.

Case $2$: $x+y=0$. This case for is trivial.

Case $3$: $x+y<0$

When $x+y<0$, we begin by making the substitution $\frac{1}{m} = -\frac{x+y}{n}$. We proceed analogously thereafter by making use of the identity

$$e^{-1}=\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n$$

Mark Viola
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  • Nice answer, just a follow up which properties of the limit do we need to pay attention to? – grayQuant Mar 19 '15 at 03:42
  • Thank you! There were no specific properties that we needed. It's just a simple substitution. Of course, we tacitly assumed $x+y>0$. But the other case can be handled similarly. – Mark Viola Mar 19 '15 at 03:49
  • Can you show $e^{-x}=(e^x)^{-1}$ – grayQuant Mar 19 '15 at 06:25
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    Note that $m$ will not be an integer after the substitution. You have to prove first that $\lim_{z\to\infty}(1+x/z)^z=e^x$ (where $z$ is supposed to be a “continuous” variable). – egreg Mar 19 '15 at 10:18
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    @egreg Where in the stated definition was $n$ restricted to be an integer? – Mark Viola Mar 19 '15 at 14:33
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    @Dr.MV: I have to agree here with egreg. When the limit is of the form $n \to \infty$ then it is implicitly assumed that $n$ is an integer unless otherwise stated. Proper answer has been provided by Jack D'Aurizio. – Paramanand Singh Mar 21 '15 at 07:31
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    @ParamanandSingh The OP believed otherwise apparently. The only requirement in the proof is that $\lim_{m \to \infty} (1+m^{-1})^m=e$. Are you suggesting that this is not true if $m$ is not an integer? If so, that belief is unfounded. – Mark Viola Mar 21 '15 at 14:30
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    @Dr.MV: the result is true even if $m$ is not an integer, but using that as a definition of $e$ is very complicated because it involves exponents which may be irrational. – Paramanand Singh Mar 21 '15 at 18:46
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    @ParamanandSingh See my answer here. http://math.stackexchange.com/questions/1198994/proof-relating-to-limit-definition-of-e/1199022#1199022 Does that seem complicated to you? And no integer restrictions! – Mark Viola Mar 21 '15 at 21:02
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    @Dr.MV: I don't want to continue this further. The question you refer to assumes that $\log x$ is defined as integral and $e^{x}$ is the inverse. This is technically the easiest approach. In the current question the starting point is the definition $e^{x} = \lim_{n \to \infty}(1 + x/n)^{n}$ and this is simpler to handle if $n$ is assumed an integer. See my posts on this topic at http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-1.html – Paramanand Singh Mar 22 '15 at 04:40
  • @ParamanandSingh I like your blog! Very nice. – Mark Viola Mar 22 '15 at 05:15
  • @ParamanandSingh You say that using that as a definition of $e$ is complicated because of irrational exponents. I don't think so, because one can get from rational exponents to irrational exponents easily by using continuity. – Poder Rac Jun 18 '20 at 21:33
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    @PoderRac: well that is rather easily said than done. Any definition of irrational exponents based on continuity and the definition of rational exponents is far complicated. By habit Rudin chooses this complicated route and defines $b^x=\sup{b^t\mid t\in\mathbb {Q}, t\leq x} $ for $b>1$. I have replaces the sup with limits and given an exposition in this blog post. – Paramanand Singh Jun 19 '20 at 00:22
  • @poderrac I suggest you read Paramanand's blog page articles. They're instructive and should facilitate the discussion you're having. – Mark Viola Jun 19 '20 at 01:55
  • @ParamanandSingh After reading your blog, I still think it's pretty intuitive, but that's probably a matter of opinion. – Poder Rac Jun 19 '20 at 20:23
  • @poderrac Mathematics is a rigorous and logical discipline, not one that relies on your intuition, which might very well be flawed. – Mark Viola Jun 19 '20 at 21:38
  • @MarkViola I think you misunderstood me, by "it" I meant the proof using continuity is pretty intuitive. – Poder Rac Jun 20 '20 at 00:16
  • @poderrac Again, intuition is not sufficient. – Mark Viola Jun 20 '20 at 01:26
  • @MarkViola I did not say that intuition is sufficient, but the proof using continuity feels very motivated and I like it. – Poder Rac Jun 20 '20 at 09:29
  • If you subtitute n by m(a+b) it must also be substituted in the limit like this: $\lim_{m(x+y)\to\infty}(1+\frac{1}{m})^{m(x+y))$ I think. – lazare Jul 07 '24 at 06:23
  • I downvote because like I said there is some wrong writing. Then you use $e^{x}$ in your proof while the grayQuant ask for only the use of the limit. Then you was validated a good aswer. – lazare Jul 08 '24 at 13:31
  • I rewrite my coment upside because I failed the LaTex and I can't edit: """If you subtitute n by m(x+y) it must also be substituted in the limit like this: $\lim_{m(x+y)\to\infty}(1+\frac{1}{m})^{m(x+y)}$ I think.""" – lazare Jul 08 '24 at 13:36
  • It change also the value of $m$ and $x+y$ for example if $m=4$ and $(x+y)\to\infty$ still $m(x+y)$ go to infinity and that is a possibility in your limit. – lazare Jul 08 '24 at 13:45
  • I admit that I voted down also because your answer was used to close the subject I opened here: https://math.stackexchange.com/questions/4941390/showing-that-fx-lim-n-to-infty-fracxn1n-has-the-property-fab but I think that it's not a fair move. I'm sorry. I will undo my downvote while the option come back. – lazare Jul 08 '24 at 13:58
  • @lazare You might be able to reverse the downvote now. – Mark Viola Jul 08 '24 at 14:03
  • It's done I think. – lazare Jul 08 '24 at 14:12
  • @lazare Yes, it is! Thank you so much for that. – Mark Viola Jul 08 '24 at 14:37