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.
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4 Answers
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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.
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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
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Hint: Consider the family of paths $y = kx^{\frac{2}{3}}$.
answered Aug 17, 2014 at 15:20
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$\begingroup$ Oh i didnt know that we can use to a power of a fraction. $\endgroup$– ys wongAug 17, 2014 at 15:21
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$\begingroup$ If the limit exists, the limit along any path exists and the limits along any two paths is the same. Note that using paths you can only show that a two-variable limit does not exist, you can't use paths to prove that a two-variable limit does exist. $\endgroup$ Aug 17, 2014 at 15:40
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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 $$y\frac{\delta q}{\delta x} - x\frac{\delta q}{\delta y} =0$$
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$\begingroup$ It is also worth mentioning that the curve y=x does not reside in the discriminantal variety of $q$. The variety is $xy^2(6 x^6 + 4 x^4 y^2 - 3 x^2 y^6 - 2 y^8)=0$, which does not contain the main diagonal. While the curve is suitable for showing this limit does not exist, it does not follow the method detailed in Marc's provided link. Using the wrong curves could lead one to erroneously conclude that a provided limit exists if one is not careful. $\endgroup$ Apr 30, 2018 at 22:04