limits of logarithm

I am trying to understand the definition of a logarithm, because when I was trying to find the derivative of $2^x$ I got $$2^x \lim_{h \to 0} \frac{2^h-1}{h}$$ which I have found by searching to be $\ln(2)$. I did get a bit confused because I would need to use l'hopital rule, which would bring be back to what I was trying to find.

But my question that I think I need to understand before getting to my second question.

Euler defines logarithm as $$\ln(x)=\lim_{n \to \infty}n(x^{\tfrac{1}{n}}-1)$$

Which then must be equal to $$\ln(x)=\lim_{h \to 0} \frac{x^h-1}{h}$$

Could you help me understand how these are both the same?

Answer

These are both the same since if you define $h = 1/n$, you get your second equation from the first.