Proving that $\lim_{h\to 0 } \frac{b^{h}-1}{h} = \ln{b}$

It is easy to see that $$\lim_{x\to 0}\frac{\log_a(1+x)}{x}=\log_a e$$ Now set $y=a^x-1$. So $a^x=1+y$ and then $$x=\log_a(1+y)$$ Clearly, when $x\to0$; $y\to 0$ so $$\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\log_e a$$

$$\text{ As }b=e^{\ln_eb}, \frac{b^h-1}h=\frac{e^{h\ln b}-1}h=\ln b\frac{e^{h\ln b}-1}{h\ln b }$$

$$\text{As, }\lim_{x\to 0}\frac{e^x-1}x=1,$$ $$\lim_{h\to0}\frac{b^h-1}h=\ln b\lim_{h\to0}\frac{e^{h\ln b}-1}{h\ln b }=\ln b$$

$$\lim_{x\to 0} \frac{\log_a(1+x)}{x} = \lim_{x\to 0} \frac{\log_e(1+x)}{x\log_e{a}}=\lim_{x\to0} \frac{\sum_{i=1}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{x\log{a}} =\lim_{x\to 0} \frac{1+\sum_{i=2}^{\infty} (-1)^{n+1} \frac{x^{n}}{n}}{\log{a}} = \frac{1}{\log{a}} = \frac{1}{\frac{\log_a{a}}{\log_{a}{e}}} = \log_{a}{e}$$

Now set $y=a^x -1.$ Hence $a^x = 1+y$ and $$ x = \log_a(1+y) $$ As $x\to 0; y\to 0$
Therefore: $$ \lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\log_e a $$

Thanks Babak.