Two variable limit problem : Polar Coordinates

[michonamona]

Homework Statement

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

Homework Equations

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?

Thank you in advanced.

M

 

[fzero]

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.

 

[michonamona]

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.

Thanks for your reply.


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

 

[fzero]

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.