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Question:

Is $$\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{nx}\right)=\lim_{n\to\infty}\left(\left(1+\frac{x}{n}\right)^{n}\right)$$

Background:

I am trying to show the equivalence of definitons of $e^x$ starting with $$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$

I proceeded as follows $$e^x=\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right)^x$$ and as the limit does not depend on $x$ then$$e^x=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$$

Next I expanded using the binomial expansion:

$$e^x=\lim_{n\to\infty}\left(1+nx\frac{1}{n}+\frac{nx\left(nx-1\right)}{2!}\left(\frac{1}{n}\right)^2+...+\right)$$

$$e^x=\lim_{n\to\infty}\left(1+x+\frac{x^2-\frac{x}{n}}{2!}+...+\right)$$ $$e^x=1+x+\frac{x^2}{2!}+...+$$ I have seen $e^x$ defined as $$\lim_{n\to\infty}\left(\left(1+\frac{x}{n}\right)^{n}\right)$$ and am trying to reconcile that fact.

Using the binomial expansion $$\lim_{n\to\infty}\left(\left(1+\frac{x}{n}\right)^{n}\right)=\lim_{n\to\infty}\left(1+n\left(\frac{x}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{x}{n}\right)^2+...+\right)$$ $$=1+x+\frac{x^2}{2!}+...+=e^x$$

So it seems the two are equivalent as $n\to\infty$ but not before. Is this correct? I feel it is and that the two functions are equivalent in the limit.

A broader question therefore is:

Are there extra algebraic manipulations such as moving the $x$ inside that are valid in limit situations? Perhaps under given constraints?

Thanks for your help in advance. I have scanned the site for the answer but didn't see anything but could have missed it.

Clayton
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Karl
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  • If you already have some definition of $\log$, you may just compare $nx\log\left(1+\frac{1}{n}\right)$ and $n\log\left(1+\frac{x}{n}\right)$. You may easily check that the difference is $O\left(\frac{|x|}{n}\right)$ as $n\to \infty$, so your claim holds by the continuity of $\exp$ and $\log$. – Jack D'Aurizio Aug 24 '18 at 18:00
  • @JackD'Aurizio Hi Jack. That reasoning seems dangerously circular if one is beginning with the limit definition of $e$. – Mark Viola Aug 24 '18 at 18:37
  • @MarkViola: well, I know, that is the reason for the initial if. On the other hand, to apply Bernoulli's inequality or to exploit the Taylor series of $\log(1+x)$ are pretty much the same thing, once $\log$ has been defined in some way. – Jack D'Aurizio Aug 24 '18 at 19:08

1 Answers1

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You can make the following change of variable $nx=k$ so $\lim\limits_{n\rightarrow \infty} \left(1+\dfrac{1}{n} \right)^{n x}=\lim\limits_{k\rightarrow \infty} \left(1+\dfrac{x}{k} \right)^k$.

fabian_mc
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  • Thank you. It seems obvious now but I'm rarely see the wood for the trees unfortunately. – Karl Aug 24 '18 at 18:22
  • @MarkViola that is a subtle comment. I simply hadn’t thought about. Is there a work around? – Karl Aug 24 '18 at 19:00
  • I unaccepted the answer. Sorry fabian it seems the right thing given Marks comment. I will accept it if there is a work around. Hope thats OK, I mean no offence. I should have thought before accepting my fault. – Karl Aug 24 '18 at 19:10
  • No problem at all, the idea of math is to be certain... So it's clear that for $x\in\mathbb{N}$ the idea works. So suppose $x\in\mathbb{Q}$, so $x=\dfrac{p}{q}$, since $\lim\limits_{n\rightarrow \infty}\left(1+\dfrac{1}{n} \right)^n$ exists then $\lim\limits_{n\rightarrow \infty}\left(1+\dfrac{1}{n} \right)^{\dfrac{n p}{q}}$ also exists, so take the subsequence $\alpha_n=n q$, so $\lim\limits_{n \rightarrow \infty}\left(1+\dfrac{1}{n} \right)^{\dfrac{\alpha_n p}{q}}=\lim\limits_{n \rightarrow \infty}\left(1+\dfrac{1}{n q} \right)^{n p}$... For $x\in\mathbb{R}$ it follows by a density argument – fabian_mc Aug 24 '18 at 20:13
  • @fabian_mc Well done. I was in the process of writing a solution. but have deleted it after my reading your comment. (+1) – Mark Viola Aug 24 '18 at 20:23