Let and be functions, both differentiable in a neighborhood of 0, with and such that
Prove or disprove each the following statements.
- as .
- as .
Proof. Since as we know by the definition of that
Thus, for every there exists a such that
So, for we have
The final line follows since by hypothesis. Therefore,
By definition, we then have
Consider for and for . Then, for ,
For we have .
Since we have as . However, since
does not exist.
Define the function:
Prove there is no such that
Proof. We prove this by contradiction. Suppose there does exist some such that . From the definition of limit this means that for all there exists a such that
First, we claim that for any such we must have . This must be the case since if then , so we may choose such that . But then, since for all , we have
(Where we used part (i) this exercise (Section I.4.9, Exercise #1) for the inequality .) This contradicts our choice of , so must be less than or equal to 1.
Next, suppose . Then choose . To obtain our contradiction we must show that there is no such that
By the Archimedean property of the real numbers we know that for any , there exists a positive integer with such that . (The Archimedean property guarantees us an integer, but then the inequality will hold for any larger integer, so we can choose one with the property of being .) But,
Then, from the definition of and since and we have,
This contradicts that if then for every we have for all . In other words, we found an greater than 0 (in particular, we found ) such that no matter how small we choose there exists an such that is smaller than , but is bigger than . This exactly contradicts the definition of . Hence, there can be no such number
This proof shows that there is now way to define so that is continuous at 0. We see this since for to be continuous at 0 we must have . But if we try to define for any , we will fail to achieve continuity since the proof shows for any we might want to choose.