Limit of $\frac{\pi^h-1}{h}$ as h approaches zero Can someone help me find this limit here. I only know how to use L'Hospital's rule but I want to be able to evaluate this limit without using differentiation.
$$\lim \limits_{h \to 0} \frac{\pi^h-1}{h}$$
The reason I want this limit is because just like $e$ can be expressed as $\sum_{n=0}^{\infty}\frac{1}{n!}$ I want to find a way to do the same with $\pi$ so i want to find the dervivative of $\pi^x$ without having $\pi$ in the result.

You don't need that limit to compute the derivative of $a^x$ though.  ($a$ a constant)
Since $a = e^{ \ln a}$ (log and exponential are inverse functions), you have $a^x = e^{(\ln a )x}$, so $$ \frac{d}{dx}(a^x) = \frac{d}{dx} e^{(\ln a)x} = (\ln a) e^{(\ln a) x}$$ 
