Does $\lim_{(x,y) \to (0,0)} xy^4 / (x^2 + y^8)$ exist? From this question on answers.yahoo, the guy says the following limit does not exist:
$$\lim_{(x,y) \to (0,0)} \frac{xy^4}{x^2 + y^8},$$
 then on wolfram, it says the limit is equal to $0$. When I did it myself, I tried approaching $(0,0)$ from the $x$-axis, $y$-axis, $y=x$, $y=x^2$. They all equal $0$.
But when I tried the squeeze theorem, I got $y^8 \leq x^2 + y^8$, therefore $0 \leq |xy^4/(x^2+y^8)| \leq |\dfrac{x}{y^4}|$, and the latter does not exist for $(x, y) \to (0,0)$.
So does the original limit exist or not? I'm getting contradicting information from various sources. Also, if it doesn't exist (it looks like it doesn't... I think), how would I go about proving that it doesn't?

Hint: let $x=ky^4(k\neq 0)$,so
$$\lim_{(x,y)\to(0,0)}\dfrac{xy^4}{x^2+y^8}=\dfrac{k}{k^2+1}$$
