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.
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