How is
$\lim_{h \to 0} \frac {3^h-1} {h}=\ln3$
evaluated?
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'(0) = \log{3}.
The result can also be obtained using $\int_a^b {e^x \,dx} = e^b - e^a$ (for all $a,b \in \mathbb{R}$). Indeed, for any $h \neq 0$ it holds $ \frac{{3^h - 1}}{h} = \frac{{\int_0^{(\ln 3)h} {e^u \,du} }}{h}, $ and hence (since $x \mapsto e^x$ is continuous and $e^0=1$) $ \mathop {\lim }\limits_{h \to 0} \frac{{3^h - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\int_0^{(\ln 3)h} {1\,du} }}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{(\ln 3)h}}{h} = \ln 3. $
If you want a philological answer, this limit must be proved only by means of the definition of $x \mapsto a^x$. Of course it is the definition of the derivative of this function at $x_0=0$, and therefore you should not use De l'Hospital's rule: how can you compute a derivative if you do not know how to compute the same derivative?
These limits are always troublesome, since most of us do not learn a rigorous definition of the exponential function before computing elementary derivatives. That's why the limit $ \lim_{x \to 0} \frac{e^x-1}{x}=1 $ is a fundamental limit: it is so hard to prove without a circular reasoning that calculus teachers take it for granted.
However, if you do not remember the final result, any tool is welcome, provided it gives you the correct answer!
This is such a basic limit that might as well be built up before the derivative, so I think the L'Hospital or derivative had better be avoided, and the following proof is necessary:
We prove that $\lim_{x\to0}\,(a^x-1)/x=\ln a$.
First we prove that $\lim_{x\to0}\,x/\log_a(1+x)=\ln a$.
$\lim_{x\to0}\frac{\log_a(1+x)}x=\lim_{x\to0}\log_a(1+x)^{1/x}=\log_ae$
(Can you prove that $\lim_{x\to0}(1+x)^{1/x}=e$ and $\log$ is continuous?)
So
$\lim_{x\to0}\frac x{\log_a(1+x)}=\ln a$
In your problem, let $y=a^x-1$, and $x\to0$ (and never $=0$), we have $y\to0$ (and never $=0$), so $(a^x-1)/x=y/\log_a(1+y)\to\ln a$.
$\lim_{h\to 0}\frac{3^h-1}{h}=\lim_{h\to 0}\frac{e^{h\log 3}-1}{h}$. Now expansion of $e^{h\log 3}=1+\frac{h\log 3}{1!}+\frac{h^2(\log 3)^2}{2!}\cdots \implies \frac{e^{h\log 3}-1}{h}=\log 3+\frac{h(\log 3)^2}{2!}\cdots \implies \lim_{h\to 0}\frac{e^{h\log 3}-1}{h}=\log 3+0+0+\cdots = \log 3$ Hence, $\lim_{h\to 0}\frac{3^h-1}{h}=\log 3$.
Using the formula $\lim_{x \to 0} \frac {a^x-1} {x}=\ln a,$ we have for $a = 3$ $\lim_{h \to 0} \frac {3^h-1} {h}=\ln 3.$