Calculating $\lim_{h \to 0}\frac{1}{h}\ln\left(\frac{e^h-1}{h}\right)$ in an elementary way I have to calculate the limit of $$\lim_{h \to 0}\dfrac{1}{h}\ln\left(\dfrac{e^h-1}{h}\right)$$
I can calculate this limit using Taylor series, and got the answer $\frac{1}{2}$. However, I want to solve this limit in a somewhat "elementary" way, not using Taylor series, Laurent series, or L'Hopital's rule. Are there any such ways?

$$\lim_{h\to0}\frac1h\ln\left(\frac{e^h-1}h\right)=\left.\frac{d}{dx}\ln\left(\frac{e^x-1}x\right)\right|_{x=0}$$
