<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] 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
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