Does the limit $\lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6}$ exist $$
\lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6} 
$$
Does this limit exist? 
I've tried substituting y=x^0.5 and y=x^(2/3) which both goes to 0.
 A: Let $A>0$ be a parameter. Consider the curve $C_A$ determined by the equation
$$
A=x^4+y^6.
$$
This curve is not at all circular - it becomes quite oblong when $A\to0$. Anyway let's use these instead of circles as the polar coordinates don't seem to lead to a convincing argument.
Because both exponents are even, on the curve $C_A$ we have $|x|\le A^{1/4}$ and $|y|\le A^{1/6}$. Therefore 
$$
|x^3y^2|\le A^{3/4+1/3}=A\cdot A^{1/12}
$$
for all points $(x,y)\in C_A$. This means that
$$
|f(x,y)|=\frac{|x^3y^2|}{x^4+y^6}\le A^{1/12}
$$
for all the points $(x,y)\in C_A$. This is all well, because $A^{1/12}\to0$ when $A\to0$.
The interiors of the curves $C_A$ actually form a basis of neighborhoods of the origin, so we could already conclude that the limit exists and is equal to zero. To make this more precise consider a disk $B((0,0);r)$ of radius $r>0$ around the origin. Inside that disk we have $x^4\le r^4$ and $y^6\le r^6$, so we see that $B((0,0);r)$ is contained in the union of the curves $C_A$ with $A$ ranging over the interval $0\le A\le r^4+r^6$.
This means that for a point $(x,y)\in B((0,0);r)$ we have the estimate
$$
|f(x,y)|\le (r^4+r^6)^{1/12}.
$$
Because
$$
\lim_{r\to0+}(r^4+r^6)^{1/12}=0
$$
the sandwich principle then implies that
$$
\lim_{(x,y)\to(0,0)}f(x,y)=0.
$$
A: We have $$   \frac{|x|^3y^2}{ x^4 + y^6 } \leq c \sqrt{ |y|}$$ 
Indeed it suffices to show that 
$$  |x|^3y^2\leq c \sqrt{ |y|} (x^4 + y^6) $$
Which we see it holds from AM - GM on $$  x^4/3 +x^4/3 +x^4/3+y^6 \geq C|x|^3\sqrt{|y|^3} $$
