>>12435134If f is continuous, then this can be proven pretty easily. Prove the contrapositive.
Because the function does not converge to 0 as x approaches infinity, you can find some such that, for all , there exist some where .
Because the function is continuous, each such that has a neighborhood around it such that all in the neighborhood satisfies .
For , we can take a closed interval , so when we find such that implies , we have it that
Taking , we find at least one neighborhood (as mentioned before) covered by . Take a closed interval subset of the neighborhood such that it has diameter , it is completely covered by , scale it by and define it to be . Note: the ai term is used to make sure that the k scale associated with the closed interval subset increases without bound.
Repeat the aforementioned process as an infinite procedure to create a nested sequence of closed intervals, with diameter approaching 0. The intersection of these intervals contains exactly one real number . Therefore, it is not true that
This completes the proof, take the contrapositive to get the corollary you want.
