1
$\begingroup$

Let $1>f(k+1,n)>f(k,n)>0$ and $0.

And $f(k,\cdot)\to1, f(\cdot,n)\to0$.

Is there an $f$ for every $0, such that $f(x,x)_{x \to \infty}\to c$?

  • 0
    Op has now edited his question, in case it goes unnoticed and this comment is noticed! =)2012-02-03

1 Answers 1

2

$f(k,n)=c^{n/k}$ 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.