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I'm trying to prove that $\lim\limits_{m \to \infty}[(1 + \frac{1}{m})^m]^x = \exp(x)$ by expanding polynomials and comparing them, given that $\exp(x)$ is already defined as:

$$\exp(x) = \sum_{n=0}^{\infty} \dfrac{x^n}{n!} = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dotsc$$

I tried going about this by firstly expanding the said limit with binomial expansion as $$\lim\limits_{m \to \infty}\left[\left(1 + \frac{1}{m}\right)^m\right]^x = \lim\limits_{m \to \infty} \left(\binom{m}{0} + \binom{m}{1} \frac{1}{m} + \binom{m}{2} \frac{1}{m^2} + \dotsb \right)^x $$

my main goal being to obtain polynomial $P(x)$ which I can compare with $\exp(x)$. But here I get stuck because, after expressing each binomial coefficient, I'm not sure what to do about the $m \to \infty$ or later on exponent $x$ for that matter.

I'd really appreciate your help.

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    $\begingroup$ Try to to find $(1+1/m)^{mx}$ first then take limit. i.e. do in the same way we do $\lim (1+1/n)^n=e$. $\endgroup$
    – PNDas
    Commented Jan 2, 2021 at 16:12
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    $\begingroup$ Has your course covered $\log$? There's a slick proof using that $\endgroup$
    – jlammy
    Commented Jan 2, 2021 at 16:47
  • $\begingroup$ @jlammy Yeah, we've covered it. But could you please write it out or at least give me an idea of what you mean by it? $\endgroup$
    – nocomment
    Commented Jan 2, 2021 at 16:52
  • $\begingroup$ You have given a definition of $\exp(x) $. The problem at hand also requires a definition of $a^b$ for $a>0$ and any real $b$. A typical definition is $\exp(b\log a) $ with some suitable definition of $\log $. If you use this definition the result follows very easily. $\endgroup$
    – Paramanand Singh
    Commented Jan 5, 2021 at 3:30

2 Answers 2

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The slick $\log$ proof I alluded to:

If you know that $\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$ for $|x|<1$, then $$\left(1+\frac{1}{m}\right)^{mx}=\exp\left(mx\log\left(1+\frac{1}{m}\right)\right)=\exp(mx[1/m+o(1/m)])=\exp(x+o(1)),$$ which implies the claim.


Here's a more mechanical proof just in case.

Substitute $m=n/x$ to get the form $a_n=\left(1+\frac{x}{n}\right)^n$. Binomial expanding implies that $(a_n)$ is monotone increasing: \begin{align*} \left(1+\frac{x}{n}\right)^n &= \sum_{k\geq0}\frac{x^k}{k!}\cdot\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right) \\ &\leq\sum_{k\geq0}\frac{x^k}{k!}\cdot\prod_{i=0}^{k-1}\left(1-\frac{i+1}{n}\right)\\ &= \left(1+\frac{x}{n+1}\right)^{n+1}. \end{align*} Then as the factors $1-\frac{i}{n}<1$, it's clear that $\exp(x)$ is an upper bound for $(a_n)$. Hence $(a_n)$ converges to some limit that is at most $\exp(x)$. Fix some $m$, then for $n\geq m$, we have $$a_n=\left(1+\frac{x}{n}\right)^n\geq\sum_{k=0}^m\frac{x^k}{k!}\cdot\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right).$$ You can see easily that for a fixed $k$, the product $\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right)$ converges to $1$ as $n\to\infty$. So taking the limit of the above inequality as $n\to\infty$ implies that $$\lim_{n\to\infty}a_n\geq\sum_{k=0}^m\frac{t^k}{k!},$$ so now taking $m\to\infty$ implies the result.

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First note $$\left(1+\frac 1 m\right)^{mx}\\=1+mx\cdot \frac 1 m+\frac{mx(mx-1)}{2}\cdot \frac 1 {m^2}+\cdots+\frac{(mx)(mx-1)\cdots(mx-n+1)}{n!}\cdot\frac 1 {m^n}+\cdots\\=1+x+\frac{x(x-\frac 1 m)m^2}{2}\cdot \frac 1 {m^2}+\cdots+\frac{x(x-\frac 1 m)\cdots(x-\frac{(n-1)}m)m^n}{n!}\cdot\frac 1 {m^n}+\cdots\\=1+x+\frac{x(x-\frac 1 m)}{2}+\cdots+\frac{x(x-\frac 1 m)\cdots(x-\frac{(n-1)}m)}{n!}\cdot+\cdots$$ Now taking limits we get $$\lim_{m\to\infty}\left(1+\frac 1 m\right)^{mx}=1+x+\frac {x^2}2+\cdots+\frac{x^n}{n!}+\cdots=e^x$$

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    $\begingroup$ The step of now taking limits needs more justification. Plus you have used the binomial theorem for general index which itself requires a lot of effort. $\endgroup$
    – Paramanand Singh
    Commented Jan 5, 2021 at 3:31
  • $\begingroup$ @ParamanandSingh, In the first line, I think you might be talking about the validity of $\lim(\sum f_n)=\sum(\lim f_n)$. But in the second line , I don't get what you are talking about, please explain. May be you are talking about the validity of binomial series or why the limit exists in the first place. Thank you. $\endgroup$
    – PNDas
    Commented Jan 5, 2021 at 5:00
  • $\begingroup$ Well the binomial theorem is a complicated result. Better to avoid it when we can. $\endgroup$
    – Paramanand Singh
    Commented Jan 5, 2021 at 7:02

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