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How to prove that the following definitions of $e^x$ are equivalent, with as simple tools as possible and without any knowledge of $e$ or logarithms ?

$$\sum_{n=0}^{\infty} \frac{x^n}{n!}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$

Also preferably, prove that this is $a^x$ for some real number $a>0$.

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You've tried using the binomial theorem on the right? – J. M. Oct 4 '11 at 16:35
A simple tool would be to note that both sides ostensibly are their own derivative and that they coincide at $0$, but then showing that these ostensible derivatives are actual derivatives would require somewhat less simple tools. – joriki Oct 4 '11 at 16:38
In your last sentence, how do you want to define $a^x$? – Chris Eagle Oct 4 '11 at 17:00
up vote 5 down vote accepted

In my answer to Combinatorial proof, I show that for $\displaystyle e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$, $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$.

To show that $\displaystyle \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$, we could simply note that $$ \left(1+\frac{x}{n}\right)^n=\left(1+\frac{1}{n/x}\right)^{(n/x)x}\tag{1} $$ and just take the limit as $n\to\infty$. However, one might complain that $n/x$ is not an integer. To calm any such complaint, consider the sandwich $$ \left(1+\frac{1}{\lfloor n/x\rfloor+1}\right)^{(\lfloor n/x\rfloor+1)x-x}\le\left(1+\frac{1}{n/x}\right)^{(n/x)x}\le\left(1+\frac{1}{\lfloor n/x\rfloor}\right)^{\lfloor n/x\rfloor x+x}\tag{2} $$ It is easy to see that both $\lfloor n/x\rfloor$ and $\lfloor n/x\rfloor+1$ are integers and that $$ \lim_{n\to\infty}\left(1+\frac{1}{\lfloor n/x\rfloor+1}\right)^{-x}=\lim_{n\to\infty}\left(1+\frac{1}{\lfloor n/x\rfloor}\right)^x=1 $$ Thus, both the left and right sides of $(2)$ tend to $e^x$. Therefore we can use $(1)$ and just take the limit as $n\to\infty$.

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Using binomial theorem: $$ \left(1+\frac{x}{n}\right)^n = \sum_{k=0}^n \binom{n}{k} \frac{x^k}{n^k} = \sum_{k=0}^n \frac{x^k}{k!} \frac{n(n-1)\cdots(n-k+1)}{n^k} = \sum_{k=0}^n \frac{x^k}{k!} \prod_{m=0}^{k-1}\left(1-\frac{m}{n}\right) $$

For each fixed $k$, the limit, as $n\to \infty$, is $\frac{x^k}{k!}$ which establishes equality of formal series.

But these both the series and the limit are known to converge for all $x$.

Also, see if the following $\lim_{n\to \infty} \left( 1 + \frac{x}{n} \right)^n = \left(\lim_{n\to \infty} \left( 1 + \frac{x}{n} \right)^{n/x} \right)^x $ for $x \not= 0$ suggests you the value for the limit.

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Don't you need to argue with uniform convergence to interchange the limits? – joriki Oct 4 '11 at 16:49
@joriki Are you talking about the first, or the last limit ? If it is the first one, then my intention was to consider formal series, not worrying about convergence, and justify the equality post-factum, since the lhs is a convergent series. Arguing by uniform convergence would have been more rigorous. – Sasha Oct 4 '11 at 17:00

I don't disagree with the other answers but I always found that these kinds of limits and interesting properties came much more naturally from what I thought the definition of $e$ really was: that $e^x$ is its own derivative.

From that starting point, let $y=e^x$, so $dy/dx = e^x$. Now, $x = \ln(y)$. And $dx/dy = 1/e^x = 1/y$. So the derivative of $\ln(x)$ is $1/x$.

By first principles, you get:

\begin{equation} \frac{d\ln(x)}{dx}=\lim_{h->0}\frac{\ln(x+h)-\ln(x)}{h}=\lim_{h->0}\frac{\ln((x+h)/x)}{h}=\lim_{h->0}\ln((1+h/x)^{1/h})=1/x \end{equation}

So swapping $x$ with $1/x$ and $h$ with $1/n$:

\begin{equation} x = \lim_{n->\infty}\ln((1+x/n)^n) \end{equation}


\begin{equation} e^x = \lim_{n->\infty}(1+x/n)^n \end{equation}

Obviously you just let $x=1$ to get a limit for $e$. For the other side, it's just the Taylor series. The series is super-differentiable and its derivative is itself. In fact, all of $e^x$'s derivatives are $1$ when $x=0$.

If you think about it, there can only be one function which is its own derivative and which goes through (0, 1). Technically, it's a first order differential equation with one initial condition. You can easily approximate it with arbitrary precision by programming or using a spreadsheet:

You could also just show that the limit on the right is its own derivative as well.

Alternate method using first principles with $e^x$:

\begin{equation} \frac{de^x}{dx}=\lim_{h->0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h->0}\frac{e^{h}-1}{h} \end{equation}


\begin{equation} \lim_{h->0}\frac{e^h-1}{h}=1 \end{equation}

Rearranging to make $e$ the subject:

\begin{equation} e=\lim_{h->0}(h+1)^{1/h}=\lim_{n->\infty}(1+\frac{1}{n})^n \end{equation}

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