# Homework Help: Limits using polar forms

1. Sep 12, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Well, I've made a double limit using the polar forms. The thing is the limit is wrong, I've made a plot, and then I saw that the limit doesn't exist, and what I wanna know is what I'm reasoning wrong, and some tips to get a deeper comprehension on this limits, and on what I am doing. For the last one I wanna know the limit value, I think it doesn't exists neither. Is it because the sine and cosine oscillates?

$$\displaystyle\lim_{(x,y) \to{(0,0)}}{\displaystyle\frac{xy}{xy+(x-y)^2}}$$

$$\begin{Bmatrix} x=r\cos\theta\\y=r\sin\theta\end{matrix}$$

$$\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{r^2\cos\theta\sin\theta}{r^2\cos\theta\sin\theta+(r\cos\theta-r\sin\theta)^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{r^2cos\theta\sin\theta}{r^2[\cos\theta\sin\theta(cos\theta-\sin\theta)^2]}}=$$

$$=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{\cos\theta\sin\theta+cos^2\theta-2\cos\theta\sin\theta+\sin^2\theta}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{r^2-\cos\theta\sin\theta}}=-1$$

r is always positive, as we defined it.

$$\displaystyle\lim_{(x,y) \to{(-1,3)}}{\displaystyle\frac{\sqrt[ ]{x+y-2}}{(x+1)^2+(y-3)^2}}$$

$$\begin{Bmatrix}x=-1+r\cos\theta\\y=3+r\sin\theta\end{matrix}$$

$$\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{-1+r\cos\theta+3+r\sin\theta}}{r^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{r}\sqrt[ ]{\cos\theta+\sin\theta}}{r^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{-1+r\cos\theta+3+r\sin\theta}}{r^2}}=\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{\sqrt[ ]{\cos\theta+\sin\theta}}{r^{\frac{3}{2}}}}$$

Bye there, thanks for posting.

Last edited: Sep 12, 2010
2. Sep 12, 2010

### jgens

I'm not going to comment on your work before this (mostly because I'm not really qualified to do so), but sin2(x)+cos2(x)=1, not r2.

3. Sep 12, 2010

### Telemachus

Right, thanks. I thought of it as beeing x and y :P

4. Sep 12, 2010

### LCKurtz

I didn't check your work but I have a question for you. If a limit in fact doesn't exist, it is usually easier to find a couple of paths along which you get different values. Have you tried that or are you using polar coordinates because it is required for the problem?

In either case I would suggest looking at the lines y = x or y = -x in either polar or rectangular form.