Multivariable limit of rational function Does the following limit exist?
$$\lim_{(x,y) \to (0,0)}\frac{x^2y^3}{x^4+(x^2+y^3)^2}$$
I tried to solve this problem using polar coordinates, but I can't simplify it. I tried the squeeze theorem, I got $0.5$, but I think this is incorrect.
 A: Let $x^2=y^3$ therefore we have
$$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^3}{x^4+(x^2+y^3)^2}=\lim\limits_{x\to0}\frac{x^4}{5x^4}=\frac{1}{5} \ \ \ (1)$$ and if we consider $x=y$ then 
$$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^3}{x^4+(x^2+y^3)^2}=0  \ \ \ \ (2)$$ so that from (1) and (2)  we see that limit doesn't exist.
A: Hint: Consider the family of paths $y = kx^{\frac{2}{3}}$.
A: I am assuming that you are considering your input rational function $q$ as a real function. In this case, the origin is an isolated zero of the denominator. Consider now the family of circles $C_r$ centered at the origin and of radius $r > $. Consider finding the minimum and maximum values of $q$ along each circle $C_r$. Applying the method of Lagrange 
multipliers implies the following relation for a point $(x, y)$ at which
$q$ reaches an optimum value on $C_r$:
$$x \frac{\partial q}{\partial x} -  y \frac{\partial q}{\partial x} = 0.$$
Note that this condition is independent of $r$. Solving this equation
over the reals yields two curves:
$$ 2 x^4 = y^6 \ \ {\rm and} \ \ y = x.$$
Following each curve towards the origin, the function $q$ converges to two different values, as suggested in Answer 1. Therefore, $q$ does not admit a limit at the origin.
This trick, based on Lagrange multiplyers, was originally proposed in 
http://www.sciencedirect.com/science/article/pii/S0747717112001204
for bivariate functions and was extended to arbitrary number of variables in
http://dl.acm.org/citation.cfm?doid=2930889.2930938
All the best,
Marc Moreno Maza
A: Although Marc
's response is a wonderful technique to implement, it contains a typo. LaGrange Multiplies put the $$\nabla (f) = \lambda \nabla (g)$$ and here Marc has choosen f(x,y) = q(x,y) and g(x,y)=c to be $x^2+y^2$ hence the system of equations is $$\frac{\delta q}{\delta x} = \lambda 2x$$
$$\frac{\delta q}{\delta y} = \lambda 2y$$ and solving for $2 \lambda$ and setting equal you arrive at $$ \frac{\delta q}{x\delta x} = \frac{\delta q}{y\delta y}$$ or correctly stated as  $$y\frac{\delta q}{\delta x} - x\frac{\delta q}{\delta y} =0$$
