If you tell me clearly, what part of my proof in post #8 and post #10, you have not understood, I will try to clarify it. Dont simply say Undefined * bounded etc
Learn from your notes that testing for continuity is not just substituting the value at that point.
Hi,
I shall try to summarize the answers to your questions. You should also read Isomorphism's posts again from the beginning.
For a given , we consider the function defined on .
First remark is that , when restricted to the intervals or , is infinitely differentiable (i.e. it can be differentiated any number of times). It is indeed obtained by product and composition of functions that are infinitely differentiable on their respective domain. And we can compute for using usual rules.
It is not clear however what happens at 0 since diverges when .
a) The continuity at means that when .
- Like you said, if , then is the product of a bounded function (the sine) by a function converging to 0 at 0, so that converges to 0 at 0. Explicitly, we have and the right-hand side converges to 0 when tends to 0. Thus is continuous.
- When , we have for . Plot this function on your calculator. When goes down from 1 to 0, goes up (continuously) from 1 to , so that oscillates infinitely many times (it is like the whole sine curve on were "compressed" (backwards) into the interval , with increasing compression rate on the neighbourhood of 0). Once you understand what the graph looks like, it is pretty obvious that won't have a limit at 0. One reason is that achieves the value 1 at points arbitrarily close to 0, whereas the continuity would mean that is close to 0 on small neighboorhoods of 0. The rigorous proof goes through a sequence like the one introduced by Isomorphism: the points such that are such that for some integer . Thus, if , we have for all , while obviously . If was continuous at 0, we would have , whereas we have : contradiction. Hence is not continuous at 0.
b) Differentiation at 0. Remember a function needs to be continuous in order to be differentiable. As a consequence, we only have to deal with the case .
is differentiable at 0 iff the ratio has a limit when , . Note that this ratio equals , and . Again, we know from the previous part that, at 0, this function converges to 0 if and it diverges if . So that is differentiable if and only if , i.e. since is an integer. In addition, in this case, we have (the value of the limit).
c) Continuity of the derivative at 0. The function needs to be differentiable, so we restrict to . First compute the derivative of on , like you usually do (Isomorphism did it). Then look at its limit at 0. The derivative is continuous iff this limit exists and is 0 (since we said that ). The derivative is found to be the sum of two terms that are pretty similar to . One of them converges to 0 for any , while the other one converges to 0 iff . Hence the sum converges to 0 iff , i.e. .
I hope this clarifies things in your mind...