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.

Ok, I there, what you do is use L'Hosptial several times:
Your limit is a limit of the type $\frac{0}{0}$. Using L'Hospital yields:
$$\lim_{x\rightarrow 0}(\ldots) = \lim_{x\rightarrow 0}\frac{\exp(2x)-\exp(x)-x\exp(x)}{(\exp(x)-1)^2}$$
Then using L'Hospital twice more (in each case you will still have limit of the type $\frac{0}{0}$) yields:
$$\lim_{x\rightarrow 0}(\ldots) = \lim_{x\rightarrow 0} \frac{4\exp{2x}-3\exp(x)-x\exp(x)}{2\exp{x}}$$
And using $\lim_{x\rightarrow 0} \exp(x) = 1$ then yiels the result of $\frac{1}{2}$ you were looking for!
Hope that helps :).
I'm reading above you can for some reason not use L'Hospital. Alternatively
you can use the Taylor expansion for the exponential function:
$$\exp(x) = 1 + x + \frac{x^2}{2} + \ldots$$
Then after rearranging the expression as the smart people above me did:
$$\frac{x\exp(x)-\exp(x)+1}{x\exp(x)-x} = \frac{x+x^2+\mathcal{O}(x^2) - 1 - x - 0.5\cdot x^2 \mathcal{O}(x^2) + 1}{x+x^2+\mathcal{O}(x^2)-x}$$
Rearranging gives:
$$\lim_{x\rightarrow 0} (\ldots) = \lim_{x\rightarrow 0} \frac{\frac{x^2}{2}+\mathcal{O}(x^2)}{x^2+\mathcal{O}(x^2)} = \frac{1}{2}$$
