What is the result of $\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}$ without L'Hôpital's rule. I have the limit
$$
\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}
$$
which I need to compute without L'Hôpital's rule.
(The result is $-\frac{1}{2}$ with L'Hôpital's rule).
Thanks.
 A: $\lim \limits_{x \to0+}$
$\frac{e^x-xe^x-1}{\left(e^x-1\right)^2}\cdot\frac{x^2}{x^2}$=
$\lim \limits_{x \to0+}$
$\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}$
$\frac{x^2}{\left(e^x-1\right)^2}$
=
$\lim\limits_{x\to 0^{+}}\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}\left[\frac{x}{\left(e^x-1\right)}\right]^2$
=$\lim\limits_{x\to 0^{+}}\frac{e^x-x-1+x-xe^x}{x^2}\cdot1$
=$\lim \limits_{x \to0+}\frac{e^x-x-1}{x^2}-\lim \limits_{x \to0+}\frac{x\left(e^x-1\right)}{x^2}$
=$\frac{1}{2}-\lim \limits_{x \to0+}\frac{e^x-1}{x}$
=$\frac{1}{2}-1=-\frac{1}{2}$
note
$\lim \limits_{x \to0+}\frac{e^x-x-1}{x^2}=\frac{1}{2}$
can be proved without Lohspital rule or series
A: We write
$$\frac {e^x-xe^x-1}{x^2}\color {red}{\frac {x^2}{(e^x-1)^2}}. $$
Let us begin by
$$L=\lim_{0^+}\frac {e^x-xe^x-1}{x^2} $$
put $x^2=t $ and
$$f (t )=e^{\sqrt {t}}-\sqrt {t}e^{\sqrt {t}}$$
then
$$L=\lim_{0^+}\frac {f (t)-f (0)}{t} =f'(0) $$
and since
$$f'(t)=-\frac {1}{2}e^{\sqrt {t}} $$
we have

$$f'(0)=-\frac {1}{2} $$

the $\color {red}{red }$ fraction goes to $1$.
the final result is $-\frac {1}{2} $.
A: It is also simple to look at Taylor expansions around $0$ up to at least second order
$$e^x = 1 + x + \frac{1}{2}x^2... $$ then $$\frac{e^x - xe^x - 1}{(e^x-1)^2}  \approx \frac{1 + x + \frac{1}{2}x^2 - x - x^2 - \frac{1}{2}x^3 - 1} {(x+ \frac{1}{2}x^2)^2} = \frac{ - \frac{1}{2}x^2  - \frac{1}{2}x^3}{x^2 + \frac{1}{4}x^4 + x^3} $$ and now by looking at the lowest order terms, the conclusion that $$\lim_{x \to 0^+} = -\frac{1}{2}$$ follows quick
A: Let's give it one more shot even though there are some good answers. We can proceed as follows
\begin{align}
L&=\lim_{x\to 0^{+}}\frac{e^{x}-xe^{x}-1}{(e^{x}-1)^{2}}\notag\\
&=\lim_{x\to 0^{+}}\frac{e^{x}-xe^{x}-1}{x^{2}}\cdot\left(\frac{x}{e^{x}-1}\right)^{2}\notag\\
&= \lim_{t\to 0^{-}}\frac{e^{-t}+te^{-t}-1}{t^{2}}\notag\\
&=-\lim_{t\to 0^{-}}\frac{e^{t}-t-1}{t^{2}}\cdot\frac{1}{e^{t}}\notag\\
&=-\frac{1}{2}\text{ (via Taylor series)} \notag
\end{align} 
