$(\epsilon ,\delta)$ Proof of $\lim_{h \to 0} \frac{a^h-1}{h}$ with a >0

Question

to find the $$\lim_{h \to 0} \frac{a^h-1}{h}$$

$ a \in \mathbb{R} $ and a >0

I know that the limit of $\lim_{h \to 0} \frac{a^h-1}{h} = ln(a) $

because $ \lim_{h \to 0} \frac{a^h-1}{h} = f'(0) $ with f(x) = $ {a^x}$ , and f'(x) = ${a^x} ln(a)$

that was my proof, but I wonder to know how to use $(\epsilon ,\delta)$ in order to proof that.

Answer 1

$$|h-0|<\delta \Rightarrow |\dfrac{a^h-1}{h}-\ln a |<\epsilon \\a^h=e^{h \ln a}=1+h \ln a +\dfrac{(h \ln a)^2}{2!}+\dfrac{(h \ln a)^3}{3!}+...\\|\dfrac{a^h-1}{h}-\ln a|=|\dfrac{(1+h \ln a +\dfrac{(h \ln a)^2}{2!}+\dfrac{(h \ln a)^3}{3!}+...)-1-h\ln a}{h}|=|\dfrac12h\ln a +o(h^2)|\\<|\dfrac12h\ln a |<\epsilon \\\to |h|<2\dfrac{\epsilon}{\ln a}\\\delta \leq 2\dfrac{\epsilon}{\ln a}$$