This limit really stamped me because i'm not allowed to use L'Hôpital's rule or Taylor's series, please help!
I think the limit is $\frac{1}{2}$, but i don't know how to prove it without the L'Hôpital's rule or Taylor's series
$$\lim_{x\rightarrow 0}{\frac{xe^x- e^x + 1}{x(e^x-1)}}$$
Best Answer
Replacing $ x $ by $\color{red}{ -x} $,
$$L=\lim_0\frac{xe^x-e^x+1}{x(e^x-1)}$$ $$=\lim_0\frac{-xe^{\color{red}{-x}}-e^{-x}+1}{-x(e^{-x}-1)}$$
$$=\lim_0\frac{-x-1+e^x}{x(e^x-1)}$$
the sum gives $$2L=\lim_0\frac{x(e^x-1)}{x(e^x-1)}=1$$ thus $$L=\frac 12$$