TITLE: Does the limit $\lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6}$ exist
QUESTION [5 upvotes]: $$
\lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6} 
$$
Does this limit exist? 
I've tried substituting y=x^0.5 and y=x^(2/3) which both goes to 0.

REPLY [4 votes]: 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 $$  x^4/3 +x^4/3 +x^4/3+y^6 \geq C|x|^3\sqrt{|y|^3} $$