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

This is based on Bernoulli's compound interest definition.

But let's say we want to find $e^x$ now.

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

Let $m = nx$ so $n = m/x$. As $n$ goes to infinity, so does $m$, so:

$$e^x=\lim_{m\to\infty}\left(1+\frac xm\right)^{m}$$

This now looks like the definition you usually see but apparently this is an invalid proof of the fact because $n$ is supposed to be an integer(??) and $m/x$ may not be.

How are you supposed to get from one to the other then?


Your argument can easily be made rigorous as follows: starting with $e= \lim_{n \to \infty} (1+\frac 1 n )^n$ we can easily show that $e= \lim_{t \to \infty} (1+\frac 1 t )^t$ where the limit is taken over the continuum; in fact, if $n \leq t< n+1$ then $(1+\frac 1 t )^t \leq (1+\frac 1 n)^{n+1} =(1+\frac 1 n)(1+\frac 1 n)^{n}$ and $(1+\frac 1 t )^t \geq (1+\frac 1 {n+1})^{n} =(1+\frac 1 n)^{-1}(1+\frac 1 n)^{n}$. Since $n \to \infty$ and $1+\frac 1 n \to 1$ as $t \to \infty$ we have proved that $e= \lim_{t \to \infty} (1+\frac 1 t )^t$. Now your argument has become valid, right?


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