I'm not sure that your original function makes that much sense; I can't say that I can tell how you got to that point.
Just checking out how normal compounded interest works, I checked Wikipedia and rediscovered the formula:
$ A=P\bigg(1+\frac{r}{n}\bigg)^{nt} $
P = principal amount (initial investment)
r = annual nominal interest rate (as a decimal)
n = number of times the interest is compounded per year
t = number of years
A = amount after time t
I've tried to break that down into good 'ole
$ A=Pe^{rt} $
but haven't had any luck; I always end up with an indeterminate value relating one and infinity that I can't break down into something manageable with L'Hopital's.--Jmason 15:53, 2 October 2008 (UTC)
This is actually in our book, but I'll redo the work here to practice with Latex and so you can see it here instead of going to the book.
$ A=P\bigg(1+\frac{r}{n}\bigg)^{nt} $
Now we we want to find the limit as n goes to infinity, since we're trying to compound the interest continuously, and therefore have an infinite number of times we compound.
$ P\lim_{n\to\infty}\bigg(1+\frac{r}{n}\bigg)^{nt} $
We now have a limit in the indeterminate form $ 1^{\infty} $
Now I'm going to stray from what the book showed us and use L'H rule to show that the limit approaches $ Pe^{rt} $
First, drop the P, we can multiply it back in later. Also, since it's in form $ 1^{\infty} $ we can try to find its limit by taking the natural log of it and its limit (Which means when we find the new limit, we have to raise e to that power to get the right limit, as you already know.)
$ \lim_{n\to\infty}\ln{\bigg(1+\frac{r}{n}\bigg)^{nt}} $
Move the nt to the front
$ \lim_{n\to\infty}nt\ln{\bigg(1+\frac{r}{n}\bigg)} $
And move the nt to the bottom by inverting it
$ \lim_{n\to\infty}\frac{\ln{\bigg(1+\frac{r}{n}\bigg)}}{\frac{1}{nt}} $
Which is now in the indeterminate form $ \frac{0}{0} $ So apply L'H rule and find derivatives of top and bottom functions:
$ \lim_{n\to\infty}\frac{\frac{-r}{(1+\frac{r}{n})n^2}}{\frac{-1}{n^2t}} $
Now the $ -n^2 $ cancel and we can take the limit as n approaches infinity.
$ \lim_{n\to\infty}\frac{rt}{1+\frac{r}{n}}=\frac{rt}{1}=rt $
Now take e to this power to get the actual limit.
$ \lim_{n\to\infty}\bigg(1+\frac{r}{n}\bigg)^{nt}=e^{rt} $
And now we simply add back in our constants to solve completely.
$ A=P\lim_{n\to\infty}\bigg(1+\frac{r}{n}\bigg)^{nt}=Pe^{rt} $
