## Thomas' Calculus 13th Edition

$e^{r}$
$\lim\limits_{k \to \infty} f(k)=e^{\lim\limits_{k \to \infty}k\ln(1+\dfrac{r}{k})}$ Re-arrange as: $e^{\lim\limits_{k \to \infty}k\ln(1+\dfrac{r}{k})}=e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{r}{k})}{1/k}}=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{kr}{k+1})}{1/k}}=e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{r}{1+1/k})}{1/k}}$ and $e^{\dfrac{\ln(1+r)}{0}}=e^{r}$