Differentiability and continuity

[metalbec]

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.
Thanks

 

[jostpuur]

How do I show that f is differentiable, but f' is discontinuous at 0?

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)=...

 

[metalbec]

well. f(x)= x^2sin(1/x). I can define f(0) to be 0.

 

[jostpuur]

well. f(x)= x^2sin(1/x). I can define f(0) to be 0.

The derivative becomes a function itself, $f':\mathbb{R}\to\mathbb{R}$. If you want to prove that it is not continuous, you should first solve its values $f'(x)$ for all x. In this case it is easiest to solve the derivative for $x=0$ and for $x\neq 0$ separately. For $x\neq 0$, you can solve $f'(x)$ by using the usual derivation rules. To solve $f'(0)$ it is best to use the definition of the derivative.

 

[metalbec]

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?

 

[jostpuur]

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?

yes! This is how you prove $f'(0)=0$.

 

[metalbec]

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?

 

[jostpuur]

If I have already established that f'(0) does not exist

You have done a mistake here. f'(0) exists.