# Thread: Limit on two variables using the polar form

1. ## Limit on two variables using the polar form

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?

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

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

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

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

r is always positive, as we defined it.

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

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

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

Bye there, thanks for posting.

2. Originally Posted by Ulysses
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?

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

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

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

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

r is always positive, as we defined it.

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

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

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

Bye there, thanks for posting.
well already at this point you see that limit doesn't exist

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

because for different values of the $\displaystyle \theta$ you have different values of limit...

3. Thanks. I've lost a sine on the way, I've already corrected it, but I think your answer holds. Anyway, as I got the same in the numerator and in the denominator for the first limit, excepting that in the denominator I got the square of the radius and the expression is negative I thought it tended to be -1 when the radius ->0 whats wrong with that?

4. Originally Posted by Ulysses
Thanks. I've lost a sine on the way, I've already corrected it, but I think your answer holds. Anyway, as I got the same in the numerator and in the denominator for the first limit, excepting that in the denominator I got the square of the radius and the expression is negative I thought it tended to be -1 when the radius ->0 whats wrong with that?
$\displaystyle \sin ^2 \theta + \cos ^2 \theta = 1$

that's something you miss there so it's not $\displaystyle r^2$ like you wrote there , so $\displaystyle \displaystyle \lim _{r \to anything}$ of something without $\displaystyle r$ is just that something

P.S. I edited a little that #2 post

5. Originally Posted by Ulysses
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?

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

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

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

$\displaystyle =\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{\ cos\theta\sin\theta+cos^2\theta-2\cos\theta\sin\theta+\sin^2\theta}}$
Okay, I am with you to here

$\displaystyle =\displaystyle\lim_{r \to{0}+}{\displaystyle\frac{cos\theta\sin\theta}{r ^2-\cos\theta\sin\theta}}=-1$
How did the "r" get back into this fraction? (Added: $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$, not $\displaystyle r^2$!)

From the line above this you can cancel both the "$\displaystyle r^2$ and $\displaystyle cos(\theta)sin(\theta)$ in the numerator with the same in the denominator getting
$\displaystyle \lim_{r\to 0^+}\frac{1}{(cos(\theta)- sin(\theta)^2}$
which clearly depends on $\displaystyle \theta$. for example, it $\displaystyle \theta= 0$, that is 1. If $\displaystyle \theta= \pi/4$ it is not defined. That alone is enough to tell you that the limit does not exist.

r is always positive, as we defined it.

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

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

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

Bye there, thanks for posting.

6. You're right hallsofIvy, what happenned with that r^2 is that I thought of the sine and the cosine as beeing x and y :P

I see my mistake now, your help was clarifying. Thank you both.

Bye there!