Monday, September 28, 2015

riemann integration - Continuity of a Function f

I've been studying different types of functions and I came across one on What is an example that a function is differentiable but derivative is not Riemann integrable, but I can't figure out why f(x)=x32sin(1x) on [0,1] is continuous, because it seems that that it doesn't exist at x=0. But I know it is differentiable on (0,1), and not Riemann integrable. Some clarification, please?

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analysis - Injection, making bijection

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