Answer
$\displaystyle\lim_{h\rightarrow 0} \dfrac{3^h-1}{h}\approx 1.0986$
Work Step by Step
We have to estimate the limit:
$\displaystyle\lim_{h\rightarrow 0} \dfrac{3^h-1}{h}$
Compute $\dfrac{3^h-1}{h}$ for values of $\theta$ close to 0:
$\dfrac{3^{-0.01}-1}{-0.01}\approx 1.0925996$
$\dfrac{3^{-0.001}-1}{-0.001}\approx 1.098009$
$\dfrac{3^{-0.0001}-1}{-0.0001}\approx 1.0985519$
$\dfrac{3^{-0.00001}-1}{-0.00001}\approx 1.0986063$
$\dfrac{3^{-0.000001}-1}{-0.000001}\approx 1.0986117$
$\dfrac{3^{-0.0000001}-1}{-0.0000001}\approx 1.0986122$
$\dfrac{3^{0.0000001}-1}{0.0000001}\approx 1.0986123$
$\dfrac{3^{0.000001}-1}{0.000001}\approx 1.0986129$
$\dfrac{3^{0.00001}-1}{0.00001}\approx 1.0986183$
$\dfrac{3^{0.0001}-1}{0.0001}\approx 1.0986726$
$\dfrac{3^{0.001}-1}{0.001}\approx 1.099216$
$\dfrac{3^{0.01}-1}{0.01}\approx 1.1046692$
Therefore we have:
$\displaystyle\lim_{h\rightarrow 0} \dfrac{3^h-1}{h}\approx 1.0986$
