Proof of Continuous compounding formula

One of the more common definitions of the constant $e$ is that: $$ e = \lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^m $$ Thus we have, with a change of variables $n = mr$, that $$ \lim_{n \to \infty} P\left(1 + \frac{r}{n}\right)^{nt}\\ = \lim_{m \to \infty} P\left(1 + \frac{1}{m}\right)^{mrt}\\ = P\left(\lim_{m \to \infty}\left(1 + \frac{1}{m}\right)^m\right)^{rt}\\ = Pe^{rt} $$ and you have your continuous compounding formula.

Let's start with the general compounding formula and take the limit as $N \to \infty$.

$$\frac{A}{P}=\lim_{n \to \infty}{\left( 1+\frac{r}{n}\right)^{nt}}$$ We start by taking logs: $$\log\left(\frac{A}{P}\right)=\log\lim_{n \to \infty}{\left( 1+\frac{r}{n}\right)^{nt}}$$ And because $log(x)$ is continuous at $x\neq0$, we can swap the limit and the log: $$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{\log\left( 1+\frac{r}{n}\right)^{nt}}$$ $$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{nt \log\left( 1+\frac{r}{n}\right)}$$ Next let's take the Taylor series of the log: $$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{nt \left( \frac{r}{n} - \frac{r^2}{2n^2} + \frac{r^3}{3n^3} - \frac{r^4}{4n^4} + \dots\right)}$$ $$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{ \left( rt - \frac{tr^2}{2n} + \frac{tr^3}{3n^2} - \frac{tr^4}{4n^3} + \dots\right)}$$

Finally let's distribute the limit operator (it's linear) and take the limits: $$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{rt} - \lim_{n \to \infty}{ \frac{tr^2}{2n} + \lim_{n \to \infty}\frac{tr^3}{3n^2} - \lim_{n \to \infty}\frac{tr^4}{4n^3} + \dots}$$ $$\log\left(\frac{A}{P}\right)= rt $$ $$\frac{A}{P}= e^{rt}$$ $$A=Pe^{rt}$$

Which was what we wanted.