Help me with this limit $$ \lim_{x\to0} {{xe^x \over e^x-1}-1 \over x}$$
I know it should equal  ${1 \over 2}$ because when i calculate with number like $0.0001$ the limit $\approx {1 \over 2}$   but i can't prove it.

Divide the top and bottom by $x$ to clean stuff up: 
$$\dots={ {{e^x \over e^x-1}-\frac{1}{x} }}\normalsize=\frac{x\cdot e^x-e^x+1}{x\cdot(e^x-1)}$$
Can you do it now?
