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0 $begingroup$ Let $displaystyle g(x, y) = frac{sin^2(x - y)}{|x| + |y|}$. Prove that $lim_{(x, y)to(0,0)} g(x) = 0$. Any help on this would be appreciated. So far I think that the fact that $|x + y| le |x| + |y|$ and $|sin(x + y)| le |x + y|$ has to be utilized but I am not sure how to do so in the epsilon delta proof. limits multivariable-calculus share | cite | improve this question edited Jun 11 '13 at 6:19 Potato 21.4k 11 89 189 asked Jun 11 '13 at 6:05 user339