Diagonal triple limit Let $1>f(k+1,n)>f(k,n)>0$ and $0<f(k,n+1)<f(k,n)<1$.
And $f(k,\cdot)\to1, f(\cdot,n)\to0$.
Is there an $f$ for every $0<c<1$, such that $f(x,x)_{x \to \infty}\to c$?

$f(k,n)=c^{n/k}$ works for the original question.
For the $f(k,\cdot)\to\infty$ version mentioned in a (now deleted) comment, $f(k,n)=2^{k-n}c^{n/k}$ works.
