<s>[INST] 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.
 [/INST] $\lim \limits_{x \to0+}$
$\frac{e^x-xe^x-1}{\left(e^x-1\right)^2}\cdot\frac{x^2}{x^2}$=
$\lim \limits_{x \to0+}$
$\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}$
$\frac{x^2}{\left(e^x-1\right)^2}$
=
$\lim\limits_{x\to 0^{+}}\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}\left[\frac{x}{\left(e^x-1\right)}\right]^2$
=$\lim\limits_{x\to 0^{+}}\frac{e^x-x-1+x-xe^x}{x^2}\cdot1$
=$\lim \limits_{x \to0+}\frac{e^x-x-1}{x^2}-\lim \limits_{x \to0+}\frac{x\left(e^x-1\right)}{x^2}$
=$\frac{1}{2}-\lim \limits_{x \to0+}\frac{e^x-1}{x}$
=$\frac{1}{2}-1=-\frac{1}{2}$
note
$\lim \limits_{x \to0+}\frac{e^x-x-1}{x^2}=\frac{1}{2}$
can be proved without Lohspital rule or series
</s>