Monday, 30 June 2014

calculus - Is f(x,y)=fracsinsqrt[3]x3+y3sqrt[5]x5+y5 uniformly continuous or not

Find out if function f(x,y)=sin3x3+y35x5+y5 is uniformly continous or not in area $D=\{0I found out that we have no lim, because \lim\limits_{x,y\to 0}\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}=\lim\limits_{\rho\to0}\frac{\sqrt[3]{\rho^3\sin^3\alpha+\rho^3\cos^3\alpha}}{\sqrt[5]{\rho^5\sin^5\alpha+\rho^5\cos^5\alpha}}=\lim\limits_{\rho\to0}\frac{\sqrt[3]{\sin^3\alpha+\cos^3\alpha}}{\sqrt[5]{\sin^5\alpha+\cos^5\alpha}}
hence we can't use The Uniform Continuity Theorem as we can't determ f(0,0). Function doesn't have bounded partial derivatives, so I think it's not uniformly continous, but I don't know how to show that

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