TITLE: Continuity of $f(x,y)=4x^3y^{11}(x^4+y^8)^{-2}$ at $(0,0)$
QUESTION [10 upvotes]: Well, the function is $$\frac{4x^3y^{11}}{(x^4+y^{8})^2}$$ and we want to know if it's continuous at $(0,0)$. I've tried as many trajectories as I could think of, and they all give $0$ as the limit. So I tried proving, by definition that its limit is in fact $0$, but to no avail.

REPLY [4 votes]: I'm adding this answer only because it brings out what is (to me) an interesting feature. First one can drop the $4$ and assume $x,y$ are positive, if one is claiming the limit is zero. Define $p(x,y)=x^3y^{11}/(x^4+y^8)^2$ and note we can split off one factor of $y$ and write $p(x,y)=y\cdot u(x,y)^2,$ where
$$u(x,y)=\frac{\sqrt{x}^3y^5}{x^4+y^8}.$$
If one can show $u$ is bounded above we're done. But $u$ may, after dividing numerator and denominator by the numerator of it, be written in terms only of the ratio $t=\sqrt{x}/y$ as $u=1/(t^5+t^{-3}).$ This has by one variable methods its maximum at $t=(3/5)^{1/8}$ which makes $u$ about $0.51604,$ so $u$ is bounded as desired.