Answer
$$\ln 3$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{\theta \to 0} \frac{{{3^\theta } - 1}}{\theta } \cr
& {\text{evaluating the limit, we get:}} \cr
& \mathop {\lim }\limits_{\theta \to 0} \frac{{{3^\theta } - 1}}{\theta } = \frac{{{3^0} - 1}}{0} \cr
& = \frac{0}{0} \cr
& {\text{the limit is }}\frac{0}{0}{\text{}}{\text{, so we apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{\theta \to 0} \frac{{d/d\theta \left( {{3^\theta } - 1} \right)}}{{d/d\theta \left( \theta \right)}} \cr
& = \mathop {\lim }\limits_{\theta \to 0} \frac{{{3^\theta }\ln 3}}{1} \cr
& = \mathop {\lim }\limits_{\theta \to 0} \left( {{3^\theta }\ln 3} \right) \cr
& {\text{evaluating the limit, we get:}} \cr
& = {3^0}\ln 3 \cr
& = \ln 3 \cr
& {\text{then}}{\text{,}} \cr
& \mathop {\lim }\limits_{\theta \to 0} \frac{{{3^\theta } - 1}}{\theta } = \ln 3 \cr} $$
