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.
There's no way we can avoid $\pi$ here. The value of this limit is $\log\pi$. Despite the method we use to get that limit, the exact value is $\log\pi$ and there's no way to avoid that. We can, however, do this without limits.



Let's say $\pi=e^{\ln \pi}$
$$\lim_{h \to 0} \frac{e^{h\ln\pi}-1}{h}$$
Now, we take @SangchiLee's advice and say $t=h\log \pi$, so that $h=\frac{t}{\log \pi}$:
$$\lim_{t \to 0} \frac{e^t-1}{\frac{t}{\log \pi}}$$
Simplify:
$$\lim_{t \to 0} \log\pi\frac{e^t-1}{t}$$
Take the $\log\pi$ out of the limit:
$$\log\pi\lim_{t \to 0} \frac{e^t-1}{t}$$
Now, the limit on the right is equal to $1$, so we have:
$$\log\pi\cdot 1=\log\pi$$
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 $$ \frac{d}{dx}(a^x) = \frac{d}{dx} e^{(\ln a)x} = (\ln a) e^{(\ln a) x}$$