# How to prove this limit exist?

How to prove $\lim_{(x,y)\to (0,0)} f(x,y)=\frac{|x|^\alpha y^4}{x^2+y^4}$ for all $\alpha>0$?

I think in order to prove this limit exists, I should the value is all the same from different direction. How to prove at here?

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$f=|x|^{\alpha} - \dfrac{|x|^{2+\alpha}}{x^2+y^4}$ consider limits for both. $|\dfrac{|x|^{2+\alpha}}{x^2+y^4}|\le \dfrac{|x|^{2+\alpha}}{x^2}$ – Yimin Feb 22 '13 at 4:21
Note $y^4/(x^2+y^4)<1$ – Maesumi Feb 22 '13 at 4:46

## 1 Answer

As Maesumi commented, we can observe that $$\frac{y^4}{x^2+y^4} \le \frac{y^4}{y^4}=1.$$ for $y\ne 0$ and if $y=0$, $\frac{y^4}{x^2+y^4} =0\le 1$. Then $$\left|\frac{|x|^\alpha y^4}{x^2+y^4}\right|\le |x|^\alpha.$$ We can easily(?) show that $$\lim_{(x,y)\to(0,0)} |x|^\alpha = 0.$$ By squeeze theorem, we are done.

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