# Two variable limit problem : Polar Coordinates

• michonamona
In summary, the original limit can be taken by taking r->0, but it is important to check for other paths that may give different limits. In some cases, switching to polar coordinates may make it easier to evaluate the limit, but it is not always necessary or appropriate.
michonamona

## Homework Statement

Find the limit of$$lim_{(x,y) \rightarrow (0,0)} xy(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})$$

## The Attempt at a Solution

We were supposed to switch to polar coordinates to solve this problem. Thus we get,

$$lim_{(r) \rightarrow (0)} rcos\theta rsin\theta (\frac{r^{2}cos^{2}\theta - r^{2}sin^{2}\theta}{r^{2}cos^{2}\theta+r^{2} sin^{2}\theta})$$

cancel out the r squares in the fraction and using double angle formulas we get$$lim_{(r) \rightarrow (0)} r^{2} cos\theta sin\theta cos 2\theta$$

I skipped a bunch of steps, I can write them out if I lose you guys.

My questions is as follows:

1.) What is the proper limit index after switching to polar coordinates, is r--->0 correct?

2.) What do we with the last line? Do we just let r approach zero and, thus, get limit = 0? or do we switch back to cartesian coordinate?

M

Last edited:
You can take the original limit by taking $$r\rightarrow 0$$. In general you have to be careful that there isn't a path to (x=0,y=0) that gives a different limit than this one. In polar coordinates this is can be the case if the result for the limit still depends on $$\theta$$ or if some $$\theta\rightarrow a$$ limit changes the $$r\rightarrow 0$$ limit.

fzero said:
You can take the original limit by taking $$r\rightarrow 0$$. In general you have to be careful that there isn't a path to (x=0,y=0) that gives a different limit than this one. In polar coordinates this is can be the case if the result for the limit still depends on $$\theta$$ or if some $$\theta\rightarrow a$$ limit changes the $$r\rightarrow 0$$ limit.

So was the notation that I used (r-->0) after switching to polar coordinates correct? I didn't understand the last sentence of your answer, would you clarify that in another way?

Thanks,
M

Taking r->0 in the way that you did is the same as taking x->0 on a path which is the line x= c y, with c nonzero. There's nothing wrong with that, but you must verify that there's no other path that could give a different limit.

For some functions, there could be another path that gives a different limit. For instance

$$\lim_{(x,y)\rightarrow 0} \frac{x}{y} = \lim_{r\rightarrow 0} \cot\theta .$$

Since this depends on $$\theta$$, the limit does not exist. In Cartesian coordinates we can see that the limit doesn't exist by comparing the limit where we first take x->0, which gives 0, with the limit along the path x=y, which gives 1.

## 1. What is a two variable limit problem in polar coordinates?

A two variable limit problem in polar coordinates involves finding the limit of a function as two variables approach a specific point in the polar coordinate system.

## 2. How do you convert a two variable limit problem from Cartesian coordinates to polar coordinates?

To convert a two variable limit problem from Cartesian coordinates to polar coordinates, you can use the following equations:

x = r cos(theta)

y = r sin(theta)

## 3. What is the difference between a one-sided and a two-sided limit in polar coordinates?

In polar coordinates, a one-sided limit involves approaching the point from only one direction, either from the positive or negative side. A two-sided limit involves approaching the point from both the positive and negative sides.

## 4. How do you determine if a two variable limit problem in polar coordinates exists?

A two variable limit problem in polar coordinates exists if the limit is the same regardless of which path is taken to approach the point. This can be determined by evaluating the limit along different paths and checking if they all give the same result.

## 5. What are some common techniques for solving two variable limit problems in polar coordinates?

Some common techniques for solving two variable limit problems in polar coordinates include converting to Cartesian coordinates, using trigonometric identities, and using L'Hopital's rule. It is also helpful to graph the function in order to visualize the limit and better understand the problem.

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