How to evaluate $\lim\limits_{h \to 0} \frac {3^h-1} {h}=\ln3$? How is

$$\lim_{h \to 0} \frac {3^h-1} {h}=\ln3$$

evaluated?
There are at least two ways of doing this: Either you can use de l'Hôpital's rule, and as I pointed out in the comments the third example on Wikipedia gives the details.

I think a better way of doing this (and Jonas seems to agree, as I saw after posting) is to write $f(h) = 3^{h} = e^{\log{3}\cdot h}$ and write the limit as $$\lim_{h \to 0} \frac{f(h) - f(0)}{h}$$ and recall the definition of a derivative. What comes out is $f&#39;(0) = \log{3}$.
Using the formula
$$\lim_{x \to 0} \frac {a^x-1} {x}=\ln a,$$
we have for $a = 3$
$$\lim_{h \to 0} \frac {3^h-1} {h}=\ln 3.$$