# Interactive Real Analysis - part of MathCS.org

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## Examples 6.5.6(b):

The function f(x) = x sin(1/x) is continuous everywhere except at x = 0, where it has a removable discontinuity. 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 We have seen before that this function has a removable discontinuity at x = 0. If we set f(0) = 0 then the function is continuous on the real line. To check differentiability, we'll have to check the limit of the difference quotient for c = 0.
Note that

(f(x) - f(0) ) / x = x sin( 1 / x) / x = sin( 1 / x )

But we have seen before that this function does not have a limit as x approaches 0. Can you recall the argument for this statement ?

Therefore, the function is not differentiable at 0. As a product and composition of functions that are differentiable for all x but zero, this function is differentiable everywhere except at x = 0.

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