# How can I prove that $\lim\limits_{m \to \infty}[(1 + \frac{1}{m})^m]^x = \exp(x)$ using polynomials?

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.

• Try to to find $(1+1/m)^{mx}$ first then take limit. i.e. do in the same way we do $\lim (1+1/n)^n=e$. Commented Jan 2, 2021 at 16:12
• Has your course covered $\log$? There's a slick proof using that Commented Jan 2, 2021 at 16:47
• @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? Commented Jan 2, 2021 at 16:52
• 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. Commented Jan 5, 2021 at 3:30

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.

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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• 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. Commented Jan 5, 2021 at 3:31
• @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. Commented Jan 5, 2021 at 5:00
• Well the binomial theorem is a complicated result. Better to avoid it when we can. Commented Jan 5, 2021 at 7:02