Answer
$e^{r}$
Work Step by Step
$\lim\limits_{k \to \infty} f(k)=e^{\lim\limits_{k \to \infty}k\ln(1+\dfrac{r}{k})}$
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 $e^{\dfrac{\ln(1+r)}{0}}=e^{r}$
