Recent content by Misirlou

Okay, so this is where I am. 0\leq\sqrt{a^{2}+b^{2}}^{2}\leq\delta^{2} So 0\leq\left|a^{2}+b^{2}\right|\leq\delta^{2} and because \left|a^{2}+b^{2}\right|\leq \left|a^{2}\right|+\left|b^{2}\right| Then 0\leq\left|a^{2}\right|+\left|b^{2}\right|\leq\delta^{2} and I would somehow slide in...

I'm not sure how I can go about that. Where should I start looking for a delta ?

Any guidance is appreciated :)

Okay, that makes more sense. So if sin^{2}(a-b)\leq sin(a-b)\leq\left|a\right|+\left|b\right| Then \left|a\right|+\left|b\right|<\left|a\right|+\left|b\right|\epsilon So 1\leq \epsilon However, I've been taught to solve for \delta first, and then prove with Given \epsilon>0 and Choose...

Homework Statement Prove with \epsilon-\delta: Lim_{(a,b)\rightarrow(0,0)}\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}=0 Hint: \left|sin(a+b)\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right| Homework Equations 0<\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}<\delta and...