# Limit $(x,y) \to (0,0)$

Prove that $\lim\limits_{ (x,y) \to (0,0) } \frac{x^2+y^2}{x^4+y^4} = \infty$.

I tried to write: $\frac{x^2+y^2}{x^4+y^4} = \frac{x^2}{x^4+y^4} + \frac{y^2}{x^4+y^4}$, and I tried to find a lower bound for $\frac{x^2}{x^4+y^4}$, but I failed, because it's not that easy to find an upper bound for $x^4+y^4$.

• Hint: write the function in polar coordinates. You're interested in the limit $r \to 0$ Nov 25 '15 at 21:26

$x^4+y^4 \leq x^4 + 2x^2y^2 + y^4 = (x^2 + y^2)^2$
If we don't have Justpassingby's nice trick, we can also do it in a more pedestrian (but also somewhat more general) way. Let $$f(x,y) = \frac{x^2+y^2}{x^4+y^4}$$ We observe that $f$ is continuous, and positive for all $(x,y)\ne 0$. It also satisfies $$f(cx,cy) = \frac1{c^2} f(x,y) \quad\text{for all }c>0$$ and these three observations are enough to conclude that it goes to $+\infty$ as $(x,y)\to (0,0)$.
Namely: the unit circle is compact and therefore the image of the unit circle under $f$ is an interval $[a,b]$ with $a>0$. Therefore the values of $f$ in the punctured disc of radius $r$ are all at least $a/r^2$, which directly leads to $f(x,y)\to +\infty$.