Let and . Prove that .
Proof. Let . For any , there is such that for , . Consider the set . If is empty, then the result immediately follows. So assume is nonempty and choose any . If , then , so which is a contradiction. Thus and . Then
or . Since is continuous, this inequality implies does not vanish at any point in .
Now assume there is a point and such that , Applying intermediate value theorem, we may assume is in the neighborhood of in . By mean value theorem, there is such that . Since cannot change its sign on , is strictly decreasing on . Especially, .
Now let , where on . Then
Since and , we have and is strictly decreasing. So must converge and we have
which contradicts that converges.
So, for all and , . This means is monotone increasing on . If for all , then must converge to some where . But this implies , leading to a contradiction. Hence for some sufficient large , and by our preceeding argument, for all . So as .