# Limit in Polar Coordinates?

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.

• What about $|r\cos\theta|$ ? :-)
– user65203
Commented Nov 29, 2020 at 16:52
• Clearly, $x+y^2=0$ is problematic.
– user65203
Commented Nov 29, 2020 at 17:16

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.
• This makes me wonder another thing: in a previous answer, which was deleted, it was suggested to use $x=r^2\cos^2(\theta)$ and $y=r\sin(\theta)$, or something like that. Can we use polar coordinates with $x=r^a\cos^{b}(\theta)$ and $x=r^c\sin^{d}(\theta)$, where $a,b,c,d \in \mathbb{R}$? I guess that the answer is no, otherwise we would get a contradiction with this limit, since it does not exist... Commented Nov 29, 2020 at 16:56