Limit of permuted numbers when each of them is approaching $0$ Let  $\sigma:N_n \rightarrow N_n$ be arbitrary permutation of $n$ numbers. Does the following limit exist?
$$\lim_{x_i \to 0} \frac{x_1+x_2^2+ \cdots x_n^n}{x_{\sigma(1)}+x_{\sigma(2)}^2+ \cdots {x_{\sigma(n)}^n}}$$
I've considered the case where $\sigma(n) = n$ for some $n \in N$ and concluded that then the limit doesn't exist but I'm having trouble figuring out when $\sigma(n) \neq n \ (\forall \ n \in N)$. I was given a hint to consider that $\sigma (1) >1$ and $\sigma(n) <n.$
The set $N_n$ I believe to be arbitrary set with $n$ elements, not necessarily natural numbers. Don't have much much more context than that but I don't think this matters for the problem.
 A: Consider $k$ with $\sigma(k)\ne k$. Then along the path $$x_i=\begin{cases}t&i=k\\t^{n+1}&i\ne k\end{cases}\qquad\text{with } t>0,$$ the numerator is between $t^k$ and $t^k+nt^{n+1}$ for $0<t<1$; in fact this is between $t^k$ and $2t^k$ for small enough $t$. Similarly, the denominator is between $t^{\sigma(k)}$ and $2t^{\sigma(k)}$ for small enough $t$. Thus the quotient is between $\frac12t^{k-\sigma(k)}$ and $2t^{k-\sigma(k)}$. If $k>\sigma(k)$ this tends $\to0$ as $t\to 0^+$; if $k<\sigma(a)$ it tends to $+\infty$ instead. But if $\sigma$ is not the identity, we can find both kind of $k$, i.e., the limit does not exist.
A: Let $x_1=t,x_2=2t.$ Then we have as required that $x_1,x_2 \to 0$ as $t \to 0,$ and
$$\frac{x_1+x_2^2}{x_2+x_1^2}=\frac{4t+1}{2t+2}\to \frac{1}{2}$$ as $t \to 0.$ So in this case, with a no-fixed point permutation, the limit happens to exist.
The same limit arises for $x_1=t,x_2=2t,x_3=3t$ when the denominator uses order $x_2,x_3,x_1,$ again a no-fixed point permutation.
