Answer
$e^{kb}$
Work Step by Step
Consider $f(x)=L=\lim\limits_{x \to \infty}(1+\dfrac{b}{x})^{kx}$
or, $\ln L=k\lim\limits_{x \to \infty} x \ln (1+\dfrac{b}{x})$
or, $\ln L=k \lim\limits_{x \to \infty} \dfrac{\ln (1+\dfrac{b}{x})}{1/x}$
Thus, $f(0)=\dfrac{0}{0}$
This shows an Inderminate form of the limit, so apply L'Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\ln L=k \lim\limits_{x \to \infty} \dfrac{bx}{x+b}=kb$
or, $L=e^{kb}$
