As told in the title, I found this equality: $$(1-p)^{-x^2} = x^2 p + \mathcal{O}(p^2)$$ and wonder whether this is true in general or whether it does only hold in the context I've seen it. It comes without further explanation.
If it is true in general, I would love to see a "proof" or an easy intuition. All I could think about was $(1-p)^{-x^2} = (1+o(1)) e^{px^2}=(1+o(1))\left(1+ \frac{px^2}{2}+\frac{p^2x^4}{6}+...\right)$, but obviously, this does not help.
Thank you very much for your help.
Almost, but not quite. Write $f(p) = (1-p)^{-x^2}$ and note that $$ f(p) = e^{-x^2 \ln(1-p)} . $$ This function is analytic around $p=0$, so we may compute it's Taylor series. The first derivative is $$ f'(p) = e^{-x^2 \ln(1-p)}\frac{d}{dp}(-x^2 \ln(1-p)) = f(p)\frac{x^2}{1-p}.$$ Furthermore, $$ f(0) = 1. $$ $$ f'(0) = x^2. $$ Thus, $$ f(p) = f(0) + f'(0)p + \mathcal{O}(p^2) = 1 + x^2 p + \mathcal{O}(p^2). $$