# Prove that specific tail distribution functions asymptotically behave like 1/n

Let $$F$$ be a cumulative distribution function such that there exists a positive and measurable function $$\beta$$ with $$\lim_{y \uparrow x^*} \frac{\overline{F}(y + x\beta(y))}{\overline{F}(y)} = \begin{cases} (1 + \xi x)^{-1/\xi}, & \xi \neq 0 \\ \exp(-x), & \xi = 0 \end{cases}$$ for every $$x \geq 0$$ satisfying $$1+\xi x > 0$$ where $$\overline{F}(x) = 1-F(x)$$ is the tail distribution function.
$$x^* = \sup\left\lbrace x: F(x) < 1 \right\rbrace$$ shall be the right corner of our distribution and let $$F$$ be continous in $$x^*$$ if $$x^*$$ is finite.

Now I want to prove that $$\lim_{n \to \infty} n\left(1-F\left(a_n\right)\right) = 1$$ holds for the pseudoinverse $$a_n = \inf \left\lbrace x: F(x) \geq 1 - \frac{1}{n} \right\rbrace$$.

It is clear that $$\limsup_{n \to \infty} n\left(1-F\left(a_n\right)\right) \leq 1$$ holds by definition of $$a_n$$. So it would be sufficient to prove the other inequality.

I know how to prove this statement for $$x < 0$$. There one has $$a_n + x\beta(a_n) < a_n$$ which leads to the statement by letting $$x$$ go to zero.