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Consider the following 2-variable function:

$$f(x,y) = \frac{x^2y}{x^4+y^2}$$

I would like to find the limit of this function as $(x,y) \rightarrow (0,0)$.

I used polar coordinates instead of solving explicitly in $\mathbb R^2 $, and it went as the following:

$$ x = r \cos \theta, \qquad y = r\sin\theta $$

Hence,

$$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4 + y^2} = \lim_{r \to 0}\frac{r^2\cos^2\theta(r\sin\theta)}{r^4\cos^4\theta + r^2\sin^2\theta}$$

This simplifies to,

$$ \lim_{r \to 0} \frac{r^3 \cos^2\theta\sin\theta}{r^2(r^2\cos^4\theta + \sin^2\theta)}$$

Simplifying $r^3/r^2$, we finally get;

$$\lim_{r \to 0} \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}$$

Now from the above, we find that as $r \to 0$ the limit is $0$.

I wanted to verify this answer so I checked on Wolfram Alpha. Explicitly without changing to polar coordinates, it said that the limit does not exist at $(0,0)$ and rightly so. Then how is it that with polar coordinates, the limit exists and is $0$? Am I doing something wrong in this method?

Also, what should I do in this situation, and when should I NOT use polar coordinates to find limits of multi-variable functions?

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1  
It's odd - this conflicts with what I was taught last semester, that converting to polar is a foolproof way of proving a limit exists. –  Emrakul Apr 15 at 7:46
    
I think @heropup's answer is more than convincing :) –  James Apr 15 at 8:10

5 Answers 5

up vote 37 down vote accepted

The limit is not defined because in order for the limit to exist, the value of the function for every possible path to $(0,0)$ must tend to the same finite value. When $y = x^2$, you have not necessarily shown that the limit is in fact $0$. When you transformed to polar coordinates and then took the limit as $r \to 0$, you are assuming that $\theta$ is a fixed constant. Therefore, you are looking only at paths that follow a straight line to the origin.

enter image description here enter image description here

Mathematica code:

F[x_, y_] := x^2 y/(x^4 + y^2)
op = ParametricPlot3D[{r Cos[t], r Sin[t], F[r Cos[t], r Sin[t]]},
     {r, 0, Sqrt[2.1]}, {t, -Pi, Pi}, PlotPoints -> 40, MaxRecursion -> 8,
     Mesh -> {10, 48}, PlotRange -> {{-1, 1}, {-1, 1}, {-1/2, 1/2}}, 
     SphericalRegion -> True, Axes -> False, Boxed -> False];
an = Show[op, ViewPoint -> {{Cos[2 Pi #], Sin[2 Pi #], 0}, {-Sin[2 Pi #], 
     Cos[2 Pi #], 0}, {0, 0, 1}}.{1.3, -2.4, 2}] & /@ (Range[40]/40);
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3  
$r$ might not depend on $\theta$, but $\theta$ is most certainly allowed to depend on $r$. For if not, then the only paths that can be considered as $r \to 0$ would be those paths that move in a straight line to $(0,0)$, and as the choice of $y = x^2$ shows, such straight-line paths do not suffice. If $\theta$ varies with $r$, the calculation of the limit is no longer necessarily $0$. –  heropup Apr 14 at 14:46
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My initial reasoning was that any value of theta will not matter since we have r multiplied with an expression containing all the theta values, and r is tending to zero. I thought that whatever there is in that expression will not matter since it will be zero regardless, because it is multiplied with r. –  James Apr 14 at 15:06
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@James If you fix $\theta$, the path you're taking to the origin is a straight line. There are many more paths than straight lines! (Great pictures, by the way!) –  Mike Miller Apr 14 at 15:53
5  
Ohhh, spinny graph! –  ja72 Apr 14 at 16:53
1  
@James If you allow θ->0 as r->0, you see that the "expression containing all the theta values"->0/0. This presents a problem and you would need to resolve it to show that f->0. –  Aaron Dufour Apr 14 at 17:56

Let $\alpha >0$, and consider the path $\gamma_\alpha(t) = (t,\alpha t^2)$. Then we have $f \circ \gamma_\alpha (t) = {\alpha t^4 \over t^4+ \alpha^2 t^4 }$, and the limit as $t \to 0$ is ${1 \over \alpha}$.

The limit exists along all of these paths, but is different. If the limit exists, its value must be independent of how $(x,y) \to 0$.

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You haven't taken into account what happens if $\theta$ is variable as a function of $r$ when $r$ goes to $0$. Choose $\theta$ so that $\sin \theta = r$, i.e. $\theta$ is approximately $r$ and you will get $\cos \theta$ is about 1 for small $r$, and then the limit will not be zero, so the limit doesn't exist.

If you want to use polar coordinates to show that a limit exists, particularly in the case where you want to show the limit is $0$ as $r \to 0$, then if you factor out a positive power of $r$ then you need to bound the remaining factor by either a constant or a multiple of a negative power of $r$ that is lower than the positive power you factored out. In your case you can't do this because when $\sin \theta = r$ you can't produce such a bound for the expression after you factor out $r$. If you had something like $r/(\cos^4 \theta + \sin^4 \theta)$ then you could bound $1/(\cos^4 \theta + \sin^4 \theta)$ by a constant for all $\theta$ and so you would then get that the limit is indeed $0$ as $r \to 0$.

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can u pls give an example as to what u r trying to say –  godonichia Nov 7 at 19:10
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I guess the problem appears if $\theta=0$ (or $\pi$), in your argument, the limit is $0$ whenever $\theta\neq 0$.

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It depends on the path you are following to travel to the " Origin " or x=0 and y=0 . If you follow the path x^2 =y ; then f(x,y) = y^2/(2*y^2) = 1/2 ...and the limit will be 1/2 .

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