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
$$\lim_{x\rightarrow 0}{\frac{xe^x- e^x + 1}{x(e^x-1)}}$$
 A: Having $$\lim\limits_{x\rightarrow 0}{\frac{xe^x- e^x + 1}{x(e^x-1)}}=
1+\lim\limits_{x\rightarrow 0}{\frac{x- e^x + 1}{x(e^x-1)}}=
1+\lim\limits_{x\rightarrow 0}{\frac{x- e^x + 1}{x^2\frac{e^x-1}{x}}}=
1-\lim\limits_{x\rightarrow 0}{\frac{e^x-1 - x}{x^2}}$$
It's only left to compute $\lim\limits_{x\rightarrow 0}{\frac{e^x-1 - x}{x^2}}$, which is not that trivial seing the answers to that question.
A: 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
$$L=\frac 12$$
A: 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 $\lim_{h\rightarrow0}\frac{e^h-1}{h}=\exp'(0)=1$.
A: First of all, let us compute $ \lim\limits_{x\to 0}{\frac{\mathrm{e}^{-x}+x-1}{x^{2}}} $:
Notice that for any $ t\in\mathbb{R} $, $ \left|\mathrm{e}^{t}-1\right|=\left|t\right|\left|\int_{0}^{1}{\mathrm{e}^{xt}\,\mathrm{d}x}\right|\leq\left|t\right|\int_{0}^{1}{\mathrm{e}^{x\left|t\right|}\,\mathrm{d}x}\leq\left|t\right|\mathrm{e}^{\left|t\right|} \cdot $
Observe that : \begin{aligned} \frac{\mathrm{e}^{-x}+x-1}{x^{2}}&=\int_{0}^{1}{\left(1-y\right)\mathrm{e}^{-xy}\,\mathrm{d}y}\\ &=\frac{1}{2}+\int_{0}^{1}{\left(1-y\right)\left(\mathrm{e}^{-xy}-1\right)\mathrm{d}y} \end{aligned}
Since $ \left|\int_{0}^{1}{\left(1-y\right)\left(\mathrm{e}^{-xy}-1\right)\mathrm{d}y}\right|\leq\int_{0}^{1}{\left(1-y\right)\left|\mathrm{e}^{-xy}-1\right|\mathrm{d}y}\leq \left|x\right|\int_{0}^{1}{y\left(1-y\right)\mathrm{e}^{\left|x\right|y}\,\mathrm{d}y}\underset{x\to 0}{\longrightarrow}0 $, we get :
$$ \frac{\mathrm{e}^{-x}+x-1}{x^{2}}\underset{x\to 0}{\longrightarrow}\frac{1}{2} $$
And thus : \begin{aligned}\lim_{x\to 0}{\frac{x\,\mathrm{e}^{x}-\mathrm{e}^{x}+1}{x\left(\mathrm{e}^{x}-1\right)}}&=\lim_{x\to 0}{\left(\frac{\mathrm{e}^{-x}+x-1}{x^{2}}\times\frac{x}{1-\mathrm{e}^{-x}}\right)}\\ &=\frac{1}{2}\times 1\\ \lim_{x\to 0}{\frac{x\,\mathrm{e}^{x}-\mathrm{e}^{x}+1}{x\left(\mathrm{e}^{x}-1\right)}}&=\frac{1}{2}\end{aligned}
