Answer
$e^r$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{k \to \infty} f(k)=e^{\lim\limits_{k \to \infty}k\ln(1+\dfrac{r}{k})}$
and $e^{\lim\limits_{k \to \infty}k\ln(1+\dfrac{r}{k})}=e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{r}{k})}{1/k}}=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{kr}{k+1})}{1/k}}=e^{\lim\limits_{k \to \infty}\dfrac{\ln(1+\dfrac{r}{1+1/k})}{1/k}}=e^{r}$