Let f(x) = x^(2n) sin(1/x) if x is not equal to 0, and let f(0)=0. Prove that f '(0),?

Question

f ''(0), f '''(0), ......, f(n) (0) exist, and that f(n) (x) is not continuous at 0.

Please show your work. Thank you.

(Note: f(n) (x) are higher order derivatives of f.)

Answer 1

if x is not = to zero then f(0) cannot be = 0
and anyway 0/0 is not equal to 0

Answer 2

f(x) is infinitely differentiable everywhere EXCEPT at x=0.

So what this question really says is that you should show the limit as x goes to 0 of f, f', f'', and so on exist and equal 0, until you get to the nth derivative.

Well, the first part is easy; just replace all the trigonometric functions by 1 and you'll have an upper bound on the absolute value of f that converges to 0. As for the nth derivative, you'll probably have the sum of a lot of terms that converge to 0 by the same logic, plus a term that is just a constant multiple of sin (1/x), which doesn't converge.

You should be able to formalize a proof from that sketch.

Good luck!