Thus your problem for m = 0 reads:
Is f(x) continuous at x = 0?
(This is a very popular problem. I am sure you would have seen it somewhere in your notes/text)
First prove the following result:
(*) exists and equal to 0 iff .
Now for part (A) we need to find m for which , Thus we need m for which . From the result (*), this happens iff .
For part (B), we need values of 'm' for which exists. But . Again by the result, the limit exists iff
Do part (C) yourself, its instructive
Ok I will do the last one too... But I will let you prove some statements on your own.
Since we have already proved that m > 1 iff f(x) is differentiable at x = 0, for continuous differentiability we need m > 1.
Now first prove that if m > 2, then the function f'(x) is continuous at 0, using (*) and making a similar rule for (prove this too!).
Now the only remaining case is m=2. For this case, .
Choose the sequence , and observe that but . Thus f(x) is not continuously differentiable at x = 0.
So at the end, let me list the exercises you have to do, to complete the proof rigorously:
(*) exists and equal to 0 iff
(***) exists and equal to 0 iff
(****) Using (**) and (***), prove that if m > 2, then is continuous at
Your sequence , but . So it is of no use.
As a side note, I am really sorry transgalactic, that I am failing in explaining this concept to you. I have already given you the answer for this in my previous posts. The sequence you can use is
I am telling the following based on our discussion.....If you take it in the right spirit, I suggest you take tutions. Sometimes a live tutor teaching a new concept is an invaluable experience.