Solving limit without L'hopital $\lim_{x\to0} \left(\frac{\frac{xe^x}{e^x-1}-1}{x}\right)$

Question

I'm trying to solve this limit using transformations & common limits but I've been unable to: $$\lim_{x\to0}\left( \frac{\frac{xe^x}{e^x-1}-1}{x}\right)$$

This limit simplifies to $1/2$ using L'Hôpital's rule but I had to use it 3 times in order to get the result. (I'm trying to prove that the function is differentiable at $x=0$.)

So if anyone is able to solve it using transformations only without L'hopital I would be grateful.

Thanks for your time and stay safe...

Answer 1

You can do it with Taylor polynomials if you take $y:=1-e^{-x}$, so the limit is$$\lim_{y\to0}\left(\frac1y+\frac{1}{\ln(1-y)}\right)=\lim_{y\to0}\frac1y\left(1-\frac{1}{1-y/2+o(y)}\right)=-\frac12.$$

Answer 2

Note

$$\lim_{x\to0}\left( \frac{\frac{xe^x}{e^x-1}-1}{x}\right) =\lim_{x\to0}\left(\frac1{1-e^{-x}}-\frac1x\right) = \lim_{x\to0}\left(\frac1x +\frac12 +O(x)-\frac1x\right)=\frac12 $$