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

Is there a formal proof of this fact without using L’Hôpital’s rule? I was thinking about using a proof
of this fact:
$$
\left.\frac{d(e^{x})}{dx}\right|_{x=x_0} = e^{x_0}\lim_{h\to 0} \frac{e^{h}-1}{h} = e^{x_0}\cdot 1=e^{x_0}
$$
that I have to help prove:
$$
\lim_{h\to 0} \frac{b^{h}-1}{h} = \ln{b}
$$
Is there a succinct proof of this limit? How does one prove this rigorously?

Solutions Collecting From Web of "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.

$$\lim_{h\to 0}\frac{b^h – 1}{h}$$

Let $x = \frac{b^h – 1}{h}$
$$h.x = b^h – 1$$
$$b^h = hx + 1$$

Applying $\ln$ both sides

$$h\ln{b} = \ln{(hx+1)}$$
$$\ln{b}=\frac{\ln{(h.x+1)}}{h}$$
$$\ln{b}=\frac{\ln{(h.x+1)}}{hx}.x$$
Now when $h\to0$;$hx\to0$;$\frac{\ln{(h.x+1)}}{hx}\to ln(e);x\to \ln(b)$

Thus we can say

$$\lim_{h\to 0}\frac{b^h – 1}{h} = \ln(b)$$