1

Let $x_1\geq ... \geq x_n = 0$ with $x = (x_1,...,x_n)$, $X =\left\{(x_1,...,x_n) \in [0,1]^n: x_1 \geq ... \geq x_n\right\},$ and $F$ be a twice continuously differentiable function with $F'(0)>0$. Evaluate the following limit: \begin{equation} \lim_{x \to 0, \ x\in X} \frac{F\left(\int_{0}^{1} \frac{(1-x_1)(1-2z)}{1-x_1 + \sum_{m=2}^n m (x_{m-1}-x_m)z^{m-1}}dz\right)}{F\left(1-\int_{0}^{1} \frac{(1-x_1)}{1-x_1 + \sum_{m=2}^n m (x_{m-1}-x_m)z^{m-1}}dz\right)}. \end{equation}

Attempt: both the numerator and the denominator converge to zero. Therefore, I apply l'Hopital's rule. However, I still get zero divided by zero. I guess that I should apply l'Hopital's rule again, but that does not seem like a good approach.

Green.H
  • 1,201
  • Wow, the first post since a long time allowing l' Hopital! You don't believe me? See here for example and the related posts. – Dietrich Burde Mar 01 '22 at 13:48
  • I have two ideas for this kind of limit. 1: this is not a single variable limit, hence L'Hopital can't be used. 2: if the limit exists, then we can simplify a lot by considering vectors $x$ satisfying $x_1=x_2= \dots = x_n$. Under this assumption you get $$\lim_{x} \frac{F(0)}{F(0)}$$ and in the case that $F(0)=0$ I think that the limit simply does not exist. – Crostul Mar 01 '22 at 16:27

0 Answers0