Answer
$-\ln 2$
Work Step by Step
Let $\lim\limits_{\theta \to 0}f(x)=\lim\limits_{\theta \to 0} \dfrac{(\dfrac{1}{2})^{ \theta}-1}{\theta}$
But $f(0)=\dfrac{0}{0}$
This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$
This implies that $\lim\limits_{\theta \to 0} \dfrac{(-\dfrac{1}{2})^{ \theta} (\ln 2)}{1}=\dfrac{(-\dfrac{1}{2})^{ 0} (\ln 2)}{1}$
Thus, $(-1) (\ln 2)=-\ln 2$
