Limit in Polar Coordinates

calculuslimitsmultivariable-calculus

EDIT: this is wrong, in fact the limit does not exist, as indicated in the answers.

Consider the following limit:
$$
\lim_{(x,y) \to (0,0)} \frac{x^2}{x + y^2}.
$$

It is easy to show, without using polar coordinates, that such a limit is $0$. Indeed,
$$
0 \leq\Bigg|\frac{x^2}{x + y^2}\Bigg| = \Bigg|\frac{x}{x + y^2}\Bigg||x| \leq |x|,
$$
and the argument follows by the squeeze theorem. Now assume that we do not know this fact and we start using polar coordinates instead. Write
$$
x = r\cos(\theta), y = r \sin(\theta).
$$
Then,
$$
\frac{x^2}{x + y^2} = \frac{\cos^2(\theta)}{\frac{\cos(\theta)}{r} + \sin^{2}(\theta)}.
$$
It is obvious that, for a fixed $\theta$, the above expression goes to zero as $r \to 0$, but of course, this is not enough to show what we want. What we have to see is that there exists a one-variable function $h = h(r)$, depending only on $r$, such that
$$
0 \leq \Bigg|\frac{\cos^2(\theta)}{\frac{\cos(\theta)}{r} + \sin^{2}(\theta)}\Bigg| \leq h(r)
$$
for every $\theta$ and $r$, and with $\lim_{r \to 0}h(r) = 0$.

The question is: how do we find the function $h = h(r)$ in our case?

EDIT: this is wrong, in fact the limit does not exist, as indicated in the answers.

Best Answer

The limit is NOT zero. It does not exist!

If $x=y^2$, as $y\to 0$, then $x\to 0$ and $$\frac{x^2}{x + y^2} =\frac{y^4}{y^2 + y^2}=y^2\to 0.$$ On the other hand, along the curve $x=-y^2+y^4$, as $y\to 0$, then $x\to 0$ and $$\frac{x^2}{x + y^2} =\frac{(-y^2+y^4)^2}{-y^2+y^4 + y^2}=\frac{y^4-2y^6+y^8}{y^4}= 1-2y^2+y^4\to 1.$$ Indeed, in your first approach, $\left|\frac{x}{x + y^2}\right|$ is NOT bounded in a neighbourhood of the origin.

Related Solutions

Limit Using Polar Coordinates – Calculus Explanation

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

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);

[Math] Proof that the limit exists using polar coordinates

This works in the particular case where the numerator and denominator of your function are both homogenous polynomials in $x$ and $y$ (and the one in the denominator has no nontrivial roots) -- that is, in every term the sum of the exponents if $x$ and $y$ is the same. In that case you have $$ \frac{p(x,y)}{q(x,y)} = r^k h(\theta) $$ where $h$ is some continuous function with period $2\pi$.

Then $h$ is necessarily bounded, and therefore you can conclude that $\lim_{x,y\to 0} \frac{p(x,y)}{q(x,y)} = 0$ if $k>0$.

However, beware that when the function doesn't split as nicely as this, note that it is not enough that the limit under $r\to 0^+$ exists separately for each $\theta$ -- even if the limit is the same for all $\theta$. For example, if $$ f(x,y) = \begin{cases} \frac{x^2+y^2}{y} & \text{when }y> 0 \\ 0 & \text{when }y\le 0 \end{cases} $$ In this case $\lim_{r\to 0^+} f(r,\theta)=0$ for each $\theta$ but $\lim_{x,y\to 0} f(x,y)$ nevertheless doesn't exist. The above reasoning breaks down because $h$ is now not continuous when $\theta$ is a multiple of $\pi$.

For another example where the function isn't even defined by cases, see Limit is found using polar coordinates but it is not supposed to exist., which Git Gud pointed out. (In that example the reasoning above breaks down even earlier; the dependence on $r$ and $\theta$ don't decouple at all).

Related Question

Calculating (multivariable) limit using squeeze thm. or polar coordinates ($\lim_{(x,y)\to(0,0) }\frac{x^2 y}{x^6 + y}$)