Answer
$A = A_0~e^{rt}$
Work Step by Step
$A = A_0~(1+\frac{r}{n})^{nt}$
Let $~~y = (1+\frac{r}{n})^{nt}$
$ln~y = ln~(1+\frac{r}{n})^{nt} = (nt)~ln~(1+\frac{r}{n}) = (t)~\frac{ln~(1+\frac{r}{n})}{1/n}$
$\lim\limits_{n \to \infty} ln~y = \lim\limits_{n \to \infty}~(t)~\frac{ln~(1+\frac{r}{n})}{1/n} = t~(\frac{0}{0})$
We can apply L'Hospital's Rule:
$\lim\limits_{n \to \infty} ln~y = \lim\limits_{n \to \infty}~(t)~\frac{\frac{-r/n^2}{1+\frac{r}{n}}}{-1/n^2}$
$\lim\limits_{n \to \infty} ln~y = \lim\limits_{n \to \infty}~(t)~\frac{r}{1+\frac{r}{n}}$
$\lim\limits_{n \to \infty}ln~y = (t)~\frac{r}{1+0}$
$\lim\limits_{n \to \infty}ln~y = rt$
$\lim\limits_{n \to \infty}y = e^{rt}$
Therefore:
$A = \lim\limits_{n \to \infty}A_0~(1+\frac{r}{n})^{nt} = A_0~e^{rt}$
