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Differentiability and continuity

  1. Mar 4, 2008 #1
    Hi. How do I show that f is differentiable, but f' is discontinuous at 0? I guess I'm just looking for a general idea to show discontinuity.
  2. jcsd
  3. Mar 4, 2008 #2
    First you will have to get clear about what f you have been given. If you are looking at some exercise, it must have given some definition f(x)=...
  4. Mar 4, 2008 #3
    well. f(x)= x^2sin(1/x). I can define f(0) to be 0.
  5. Mar 4, 2008 #4
    The derivative becomes a function itself, [itex]f':\mathbb{R}\to\mathbb{R}[/itex]. If you want to prove that it is not continuous, you should first solve its values [itex]f'(x)[/itex] for all x. In this case it is easiest to solve the derivative for [itex]x=0[/itex] and for [itex]x\neq 0[/itex] separately. For [itex]x\neq 0[/itex], you can solve [itex]f'(x)[/itex] by using the usual derivation rules. To solve [itex]f'(0)[/itex] it is best to use the definition of the derivative.
  6. Mar 4, 2008 #5
    Would I use the Pinching Theorem to show that because the limit as h approaches 0 from both sides is 0, then the limit of hsin (1/h) is also 0?
  7. Mar 4, 2008 #6
    yes! This is how you prove [itex]f'(0)=0[/itex].
  8. Mar 4, 2008 #7
    Okay, I think I've done that. But I'm having a hard time grasping it. If I have already established that f'(0) does not exist, how did I just show that it is zero? Or does the fact that its limit is zero in itself show that it is discontinuous at 0?
  9. Mar 4, 2008 #8
    You have done a mistake here. f'(0) exists.
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