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?

$\lim_{h\to 0}\frac{3^h-1}{h}=\lim_{h\to 0}\frac{e^{h\log 3}-1}{h}$. 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, $\lim_{h\to 0}\frac{3^h-1}{h}=\log 3$.
