Finding $\lim_{(x,y) \to (0,0)} \frac{x^5y^5}{|x|^9 + |y|^{11}}$ Setting $y = mx^k$ suggests the limit evaluates to $0$, and as far as the techniques I know, dismisses the possibility of its non-existence. But I'm having trouble using the Squeeze Theorem to prove that it does exist. 
$$\lim_{(x,y) \to (0,0)} \frac{x^5y^5}{|x|^9 + |y|^{11}}$$
Major Edit: $(x,y) \to (0,0)$ not $(x,y) \to \infty$
 A: Let $u=x^9$ and $v=y^{11}$. Then
$$
\left| \frac{x^5 y^5}{|x|^9+|y|^{11}}\right|
=\frac{\left|u^{\frac{5}{9}}v^{\frac{5}{11}}\right|}{|u|+|v|}
$$
If $|u| \geq |v|,$ then
$$
\frac{\left|u^{\frac{5}{9}}v^{\frac{5}{11}}\right|}{|u|+|v|}
\leq \frac{\left|u^{\frac{5}{9}}u^{\frac{5}{11}}\right|}{|u|}
=|u|^{\frac{1}{99}}
=\max \left\{ |u|^{\frac{1}{99}} , |v|^{\frac{1}{99}} \right\}
$$
If $|v| \geq |u|,$ then
$$
\frac{\left|u^{\frac{5}{9}}v^{\frac{5}{11}}\right|}{|u|+|v|}
\leq \frac{\left|v^{\frac{5}{9}}v^{\frac{5}{11}}\right|}{|v|}
=|v|^{\frac{1}{99}}
=\max \left\{ |u|^{\frac{1}{99}} , |v|^{\frac{1}{99}} \right\}
$$
Therefore, the following always holds:
$$
\left| \frac{x^5 y^5}{|x|^9+|y|^{11}}\right|
\leq
\max \left\{ |u|^{\frac{1}{99}} , |v|^{\frac{1}{99}} \right\}
=
\max \left\{ |x|^{\frac{1}{11}} , |y|^{\frac{1}{9}} \right\}
$$
which can be used for the Squeeze Theorem.
A: I find it clarifies things to use Lagrange Multipliers on the ratio. In this case, because of the absolute sign, we stick to the first quadrant. Multipliers says take
$$  x = A t^3 \; , \; \; y = B t^2   $$
so the fraction gives
$$ \frac{A^5 t^{15}  B^5 t^{10}}{A^9 t^{27} + B^{11}t^{22}} = \frac{C t^{25}}{D t^{22} + E t^{27}}= \frac{C t^3}{D + E t^5}$$  and does give zero.
A: By weighted AM-GM inequality, we get
\begin{align*}
\lvert x \rvert^9 + \lvert y \rvert^{11}
&= \frac{11}{20} \cdot \frac{20}{11} \lvert x \rvert^9 + \frac{9}{20} \cdot \frac{20}{9} \lvert y \rvert^{11} \\
&\geq \left(\frac{20}{11} \lvert x \rvert^9\right)^{11/20} \left(\frac{20}{9} \lvert y \rvert^{11}\right)^{9/20} \\
&= \text{const} \cdot \lvert x \rvert^{99/20} \lvert y \rvert^{99/20}.
\end{align*}
Hence
\begin{align*}
\frac{\lvert x \rvert^5 \lvert y \rvert^5}{\lvert x \rvert^9 + \lvert y \rvert^{11}}
\leq \text{const} \cdot \lvert x \rvert^{1/20} \lvert y \rvert^{1/20},
\end{align*}
which shows that
$$
\lim_{(x, y) \to (0, 0)} \frac{x^5 y^5}{\lvert x \rvert^9 + \lvert y \rvert^{11}} = 0.
$$
