Compound interest formula and continuously compounded interest formula derivation My textbook gives the formula for compound interest as: 
$A\left( t\right) =P\left( 1+\dfrac {r}{n}\right) ^{nt}$
Where:
P = The principal, r=the annual rate of interest, n= the frequency of compounding, t=Time in years and A is the total interest accrued over time.
It then goes onto show how if we compound £1 continuously at a rate of 100% for 1 year, for greater and greater values of $n$ we get:
$\left( 1+\dfrac {1}{n}\right) ^{n}\rightarrow e$
And then uses this to derive the formula for continuously compounded interest:
$A\left( t\right) =Pe^{rt}$
The book says it uses "a little calculus and the definition of e" to derive this, but how exactly does it do this?
 A: Let's see. The limit claim is pretty widely discussed on MSE, so I'm assuming you're willing to believe that the limit does approach $e$. Once you have that, you can look at the formula
$$
A_n(t) = P \left(1+ \frac{r}{n}\right)^{nt}
$$
and do a little fiddling. Let $m = \dfrac{n}{r}$, so that $n = rm$. Then rewrite:
$$
A_n(t) = P \left(1+ \frac{r}{n}\right)^{nt} = P \left(1 + \frac{1}{m}\right)^{mrt}= P \left(\left(1 + \frac{1}{m}\right)^m \right)^{rt}
$$
Now as $n \to \infty$, the thing inside the large parentheses approaches $e$, so you get 
$$
A(t) = Pe^{rt}.
$$
As for the main limit, the usual approach is to say you want to find 
$$ 
L = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n
$$
but instead, you compute 
\begin{align}
\ln L 
&= \lim_{n \to \infty} \ln \left(1 + \frac{1}{n} \right)^n \\
&= \lim_{n \to \infty}  n \ln \left(1 + \frac{1}{n} \right) \\
&= \lim_{n \to \infty}  \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n},
\end{align}
which you can evaluate with L'hopital's rule (take derivative of top and bottom, since both 
go towards 0):
\begin{align}
\ln L 
&= \lim_{n \to \infty}  \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n} \\
&= \lim_{n \to \infty}  \frac{\frac{1}{1 + \frac{1}{n}}\left(\frac{-1}{n^2}\right)} {\frac{-1}{n^2}}\\
&= \lim_{n \to \infty}  \frac{1}{1 + \frac{1}{n}}\\
&= 1.
\end{align}
Since the natural log of your limit is $1$, the limit itself must be $e$.$$$$$$   
A: There is an easier way to derive (without Calculus), using $(1 + \frac 1 m)^m = {\rm e }$.
If $A = P (1 + \frac r n)^{nt}$, then $A = P \left( 1 + \frac 1 {\frac n r} \right)^{nt}$ (we basically just took the reciprocal of $\frac r n$ and put a $1$ on top again to make it equivalent.)
Now set $\frac n r = m$ and substitute: $A = P (1 + \frac 1 m)^{nt}$. It seems pretty close now, doesn't it!
Let's reform the $\frac n r = m$ statement, shall we? We can change that into $n = mr$; substitute that back into the $nt$ exponent part for $n$, and you will get $A = P (1 + \frac 1 m)^{mrt}$ (hey, that looks kind of familiar, doesn't it!).
Substitute back in from the beginning formula $(1 + \frac 1 m)^m = {\rm e}$ and you get $A = P({\rm e})^{rt}$.
