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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 $x^4+y^4$.

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    $\begingroup$ Hint: write the function in polar coordinates. You're interested in the limit $r \to 0$ $\endgroup$ Nov 25 '15 at 21:26
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$x^4+y^4 \leq x^4 + 2x^2y^2 + y^4 = (x^2 + y^2)^2$

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  • $\begingroup$ Yep! That is the trick. +1 $\endgroup$
    – Mark Viola
    Nov 25 '15 at 21:33
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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 $f(x,y)\to +\infty$.

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