Define h : R - R by
h(x) = {x^2, x rational
h(x) = {0, x irrational
1. Prove that h is dierentiable at 0.
2. Prove that h is not continuous at c not = 0. You may use the fact that any
interval in R contains both rational and irrational points.
(Hint: Split the proof into two cases: one for c rational, the other for c irrational.)
3. Prove that h' is a function whose domain is {0} and that h" does not
exist.
Do as they tell you: split in tow cases, rational and irrational: if then there exists a sequence of rational points that --> c, but ALSO a sequence of irrational points that --> c . If f were continuous at c then f(x) --> f(c) no matter how we choose to make x --> c, but over rationals and over irrationals we get two different results.
Tonio
Tonio was assuming that you knew that a function is differentiable at x= 0 if and only if exists. And that will be true if and only if exists and is the same for all sequences that converge to 0. Since a sequence will converge as long as subsequences converge, it is sufficient to consider sequences consistin only of rational numbers and only of irrational numbers.