Computing $\lim_{x\rightarrow 0}{\frac{xe^x- e^x + 1}{x(e^x-1)}}$ without L'Hôpital's rule or Taylor series 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
$$\lim_{x\rightarrow 0}{\frac{xe^x- e^x + 1}{x(e^x-1)}}$$

How about using the Cauchy's mean value theorem (L'Hospital rule can be seen as a specialization of this). Let $f(x)=xe^x-e^x+1$ and $g(x)=xe^x-x$, then $f(0)=0=g(0)$ and by the (generalize) mean value theorem, there is $c_x$ between $0$ and $x$ such that
$$f'(c_x)(g(x)-g(0))=g'(c_x)(f(x)-f(0)).$$ This can be expressed as
$$
\frac{f(x)}{g(x)}=\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(c_x)}{g'(c_x)}=\frac{c_xe^{c_x}}{c_xe^{c_x}+ e^{c_x}-1}=\frac{e^{c_x}}{e^{c_x}+\frac{e^{c_x}-1}{c_x}}$$
As $x\rightarrow 0$, $c_x\rightarrow 0$ and so
$$\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow0}\frac{e^{c_x}}{e^{c_x}+\frac{e^{c_x}-1}{c_x}}=\frac{1}{2}$$
Here we have use the fact that $\lim_{h\rightarrow0}\frac{e^h-1}{h}=\exp'(0)=1$.
