- #26

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then as h goes to zero, only 0^0=1 remains, giving ln(x)$$\dfrac{x^h-1}{h}=\dfrac{\exp(h\log(x))-1}{h}=\dfrac{1}{h}\left(\sum_{k=0}^\infty \dfrac{h^k\log^k(x)}{k!}-1\right)=\sum_{k=0}^\infty \dfrac{h^k\log^{k+1}(x)}{(k+1)!}$$

nice.

But are you sure this is the same limit as mine? they converge at different rates...