I am trying to determine and prove the limit for the function $f(x,y)=\frac{y^2-4|y|-2|x|}{|x|+2|y|}$ after setting $y=mx$ I found out that the limit is not dependent on m and -2 is a potential candidate for the limit so I set up my equation for the squeeze theorem $\lim_{(x,y)\to(0,0)} |\frac{y^2-4|y|-2|x|}{|x|+2|y|}+2|=\lim_{(x,y)\to(0,0)}\frac{y^2}{|x|+2|y|}$ and from this point on I'm having a bit of trouble to set up an inequality to satisfy the squeeze theorem. Would I be able to claim that $y^2 \leq y^2(|x|+2|y|)$
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I think if you take $z=\text{max}(|x|,|y|)$ then $z<\delta$ and $$\dfrac{|y|^2}{|x|+2|y|}\leq\frac{z^2}{z}=z<\delta$$

Mikasa
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