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**twizter** A function is analytic in an open disc centered at b (in this case an open interval) if and only if its Taylor series converges to the value of the function at each point of the disc. Do you see the if and only if statement there? It means that you can use one to prove/disprove the other. So now look at the point x = 1. The function is not analytic at that point. So if |x| >= 1, you have a non-analytic point in your open interval which implies that the Taylor series expanded about 0 does not converge to the value of the function sqrt(1-x) at that x. But for |x| < 1, the function is analytic in the open interval (-y,y) (where is |x| < y < 1) and so at every point in this interval the series converges to the function value.