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- Thread starter solakis
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- Feb 7, 2012

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Given A>0, find a B>0 such that :

for all x,y,a,b : if 0<x<1, 0<y<1, 0<a<1, 0<b<1, and |x-a|<B, |y-b|<B, then \(\displaystyle |x^2y-a^2b|<A\)

Next (using the triangle inequality), $|x^2y - a^2b| = |x^2y-a^2y + a^2y - a^2b| \leqslant |x^2y-a^2y| + |a^2y - a^2b| = |x^2-a^2|y + a^2|y-b| \leqslant 2|x-a| + |y-b| < 3B.$

So take $B = \frac13A$. Then $|x^2y - a^2b| < 3B = A$.

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