Answer
$-\ln 2$
Work Step by Step
Consider: $\lim\limits_{\theta \to 0}f(x)=\lim\limits_{\theta \to 0} \dfrac{(\dfrac{1}{2})^{ \theta}-1}{\theta}$
We need to check that the limit has an indeterminate form.
Thus, $f(0)=\dfrac{0}{0}$
The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
Then
$\lim\limits_{\theta \to 0} \dfrac{(-\dfrac{1}{2})^{ \theta} \ln 2}{1}=\dfrac{(-\dfrac{1}{2})^{ 0} \ln 2}{1}=-\ln 2$