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Examples 6.5.6(c):

The function f(x) = x2 sin(1 / x ) has a removable discontinuity at x = 0. If the function is extended appropriately to be continuous at x = 0, is it then differentiable at x = 0 ?
 Your browser can not handle Java applets To change the function into a continuous function, we set f(0) = 0 This function is now differentiable at 0, because:
• (f(x) - f(0)) / (x - 0) = x sin( 1 / x )

Since | x sin( 1 / x ) | < | x | for all x, we see that the limit of the difference quotient for c = 0 equals zero. Hence, f is differentiable at 0, and

• f'(0) = 0

The function is also differentiable everywhere else, since it is the product and composition of differentiable functions everywhere but for x = 0. Therefore, the function is differentiable on the whole real line.

Actually, this function is more interesting than it seems, because it is

• continuous
• once differentiable
• derivative is not continuous

Thus, it provides an example to show that even if derivatives exist, they do not necessarily have to be continuous. These statements are proved at a later point, but you might want to try it on your own already.

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