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## Homework Statement

Prove with [itex]\epsilon-\delta[/itex]: [itex]Lim_{(a,b)\rightarrow(0,0)}\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}=0[/itex]

Hint: [itex]\left|sin(a+b)\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|[/itex]

## Homework Equations

[itex]0<\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}<\delta[/itex]

and

[itex]\left|f(x,y)-L\right|<\epsilon[/itex]

## The Attempt at a Solution

I tried the polar conversion, which was just messy and got me nowhere. Then I tried inputting for [itex]\epsilon-\delta[/itex] formula: [itex]0<\sqrt{x^{2}+y^{2}}<\delta[/itex] and [itex]\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}-0<\epsilon[/itex] so that [itex]sin^{2}(a-b)<(\left|a\right|+\left|b\right|)\epsilon[/itex]

The tutorials I've seen use the [itex]\epsilon[/itex] inequality and transform it into the [itex]\delta[/itex] inequality, but I don't know how to do this. I'm just looking for a way to solve it, not necessarily using the "hint".

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