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

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

i know that the limt 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 .

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

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    ah thanks , the trick was then about ${e^x} = \sum_{k=0}^{\infty} \frac{x^k}{k!}$ thank for reminding me – 2017-02-04