I don't understand where is the flaw in my logic, or where my mistake lies. It would appear that either the squeeze theorem is wrong or that it has some limitations in showing that a limit does exist in three dimensional space (implying I didn't make any mistakes).
If I have a function $f$ such that: $$ f(x,y) = \frac{xy^{4}}{x^{2}+y^{8}},\space \space (x,y)\not=0 \\ $$
And I want to analyze the limit at (0,0)
It can be solved by the squeeze theorem, since for any value of $y$: $y^{8}\ge y^{2}\ge 0$
And as of $x$: $\vert x\vert \ge 0$ and $x^{2}\ge 0$
So: $x^{2}+y^{8} \ge y^{4}\ge 0$
Then: $1 \ge \frac{y^{4}}{x^{2}+y^{8}} \ge 0$
And I can multiply by $\vert x\vert$ to finally find that $lim_{(x,y)\to (0,0)}{\frac{\vert x\vert y^{4}}{x^{2}+y^{8}}}$
Because all of the terms except $\vert x \vert $ are positive then $\lim_{(x,y)\to (0,0)}{\frac{\vert x\vert y^{4}}{x^{2}+y^{8}}}$ = $\lim_{(x,y)\to (0,0)}{\vert \frac{xy^{4}}{x^{2}+y^{8}}\vert}$
Which should be sufficient to show that the limit of $\lim_{(x,y)\to (0,0)}{f(x,y)}=0$
However! If I were to use the 'paths' method, I get a different answer. If I let $x=y^4$ then:
$$\lim_{(y)\to (0)}{\frac{y^{8}}{2y^{8}}}=\frac{1}{2}$$
And for $x=0$:
$$\lim_{(y)\to (0)}{\frac{0(y^{8})}{0+y^{4}}}=0$$
Which shows that the limit does not exist.
So, does the limit exist or does it not? If it does exist, where is the flaw in my process for showing it's existence? If it doesn't exist, what are the limitations of the squeeze theorem for showing that a limit does not exist?
$y^8 \geq y^2 $ isn't correct for $|y| < 1$. For example, $(\frac12)^8 = \frac1{256} < (\frac12)^2 = \frac14$.