Proof of Continuous compounding formula Following is the formula to calculate continuous compounding
A = P e^(RT)
Continuous Compound Interest Formula
    where,  P = principal amount (initial investment)
r = annual interest rate (as a decimal)
t = number of years
A = amount after time t

The above is specific to continuous compounding. The general compounding formula is
$$A=P\left(1+\frac{r}{n}\right)^{nt}$$
I want to understand how continuous compounding formula is derived from general compounding formula, given t=1, n=INFINITY. 
 A: 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.
A: 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. 
A: Rewrite your formula as $$ A=P \left[ \left( 1+ \frac{r}{n} \right)^{\frac{n}{r}} \right]^{rt} $$ and let $ w=\frac{n}{r} $ then
$$ A= P\left[ \left( 1+\frac{1}{w} \right)^w \right]^{rt} $$
As the frequency of compounding $n$ is increased the newly created $w$ will increase as well; thus as $ n \rightarrow \infty $ , $w \rightarrow \infty $ as well and the bracketed expression tends to the number $e$. 
Consequently we find that as $ n \rightarrow \infty $ 
$$ A = Pe^{rt} $$
