TITLE: How to evaluate $\lim\limits_{h \to 0} \frac {3^h-1} {h}=\ln3$?
QUESTION [8 upvotes]: How is
$$\lim_{h \to 0} \frac {3^h-1} {h}=\ln3$$
evaluated?

REPLY [2 votes]: 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 $(a^x-1)/x=y/\log_a(1+y)\to\ln a$.