Need some help in showing that the function f '(x) is not continuous at 0.

Here is the problem: Consider the function f(x) defined by

f(x)= x^{2}sin(1/x) if x does not = 0

0 if x = 0

Show that the function f ' (x) is not continuous at 0.

Any help would be appreciated.

Re: Need some help in showing that the function f '(x) is not continuous at 0.

Quote:

Originally Posted by

**marsalanuddin** Here is the problem: Consider the function f(x) defined by

f(x)= x^{2}sin(1/x) if x does not = 0

0 if x = 0

Show that the function f ' (x) is not continuous at 0.

Any help would be appreciated.

Use the limit definition of the derivative and the squeeze theorem to prove that $\displaystyle f(x)$ is differentiable. Then just use the product and chain rule to find the derivative. Once you have $\displaystyle f'(x)$ show that $\displaystyle \lim_{x \to 0}f'(x) \ne f'(0)$

Hint the limit of the derivative does not exist