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When working on limits of functions with two variables, $f(x,y)$, I like to convert the problem to polar coordinates a lot of the time, by changing the question from $$\lim_{(x,y)\to (0,0)}f(x,y)$$ to $$\displaystyle\lim_{r\to 0}f(r\cos\theta,r\sin\theta).$$ I was just doing some problems in my book when I encountered a limit of a function with three variables, $f(x,y,z)$. I was just wondering if there was a way to calculate such a limit with polar coordinates.

An example being: $$\lim_{(x,y,z)\to(0,0,0)}\frac{xy+yz^2+xz^2}{x^2+y^2+z^4}$$

Converting it into polar coordinates gives me:

$\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta+r\sin\theta \cdot z^2+r\cos\theta\cdot z^2}{r^2(\sin^2\theta+\cos^2\theta)+z^4}=\displaystyle\lim_{r\to 0}\dfrac{r(r\sin\theta\cos\theta+\sin\theta\cdot z^2+\cos\theta\cdot z^2)}{r^2+z^4}$

Can I proceed or is polar coordinates strictly for use with two variables only?

gebruiker
  • 6,272
Sujaan Kunalan
  • 11,194

2 Answers2

2

By substituting $x=r\cos\theta, y=r\sin\theta$ in the formula $f(x,y,z)$, you are not converting to "polar coordinates". A polar coordinate system is a two dimensional coordinate system by definition of the term.

Then what are you doing?
Well, the conversion you made, yields a system of coordinates that is known as a cylindrical coordinate system.

Why do we convert to polar coodinates sometimes?
Because $(x,y)\to (0,0)\iff r\to 0$, assuming the canonical conversion. This can make things easier, because now we only have to consider one variable $r$ in stead of two variables $x$ and $y$. However, mind that $\lim_{r\to0}$ needs to be treated with care. See this, this and this for instance.

Did I do something wrong?
Well, not yet. The substitution you made isn't wrong, is just not necessarily useful. If you convert to cylindrical coordinates and let $r\to0$, then you are not approaching the point $(0,0)$ but the $z$-axis. So if you were to continue using this method, you would have to calculate $\lim_{(r,z)\to(0,0)}$ (also a tricky thing). Because only then are you approaching $(0,0)$.

Is there a three dimensional equivalent of the polar coordinate system?
Yes, there is. It's called the spherical coordinate system. Once you've converted from cartesian coordinates to spherical coordinates, we have that $(x,y,z)\to(0,0,0) \iff r\to 0$. Once again, it will suffice to consider only one variable $r$ now, if we're lucky (mind that $\lim_{r\to0}$ is still a tricky thing).

I did not perform any calculations on your limit. I'm leaving that as an exercise to you. I would like to leave you with the note that converting to spherical coordinate isn't a magic way to solve any $\lim_{(x,y,z)\to(0,0,0)}f(x,y,z)$ problem, The same holds for polar coordinates, or any coordinate transformation for that matter. If it works, it works. If it doesn't, too bad.

gebruiker
  • 6,272
0

The easy way to show that a limit does not exist is to show that we get two different answers approaching (0, 0, 0) in different directions. In each case, let u -> 0 Case 1 $x = u$; $y = u$; $z=\sqrt u$ Substituting, we get $3u^2/3u^2 = 1$ Case 2 $x = u$; $y = -u$; $z = \sqrt u$ Substituting, we get $u^2/3u^2 = 1/3$ Two different limits, so no limit.

lennon310
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T. K.
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