Also die Aufgabe muss lauten
$$ \lim_{h\to 0} a^h\frac{a^h-1}{h}=\lim_{h\to 0} a^h \frac{e^{h\cdot ln(a)}-1}{h}=\lim_{h\to 0} a^h \left[ \frac{ \sum_{k=0}^{\infty} \frac{h^k\cdot (ln(a))^k}{k!} - 1}{h} \right] = $$
$$\lim_{h\to 0} a^h \frac{ \sum_{k=1}^{\infty} \frac{h^k\cdot (ln(a))^k}{k!} }{h} = \lim_{h\to 0} a^h \sum_{k=0}^{\infty} \frac{h^{k}\cdot (ln(a))^{k+1}}{(k+1)!} = \lim_{h\to 0} a^h \left[ ln(a)+\sum_{k=1}^{\infty} \frac{h^{k}\cdot (ln(a))^{k+1}}{(k+1)!} \right] = ln(a) $$
