# Thread: Continouty of a mapping, need help

1. ## Continouty of a mapping, need help

Hi,

I'm having trouble with an assignment of mine. I hope someone here can help me. It should be trivial - I just think I've overseen something... Well, here goes:

I have a mapping $\displaystyle \theta : \mathbb{R}^2 \setminus \{ (x,y) | y = 0, x \leq 0 \} \rightarrow (-\pi,\pi)$ defined such that $\displaystyle \frac{x}{\sqrt{x^2 + y^2}} = \cos\theta(x,y)$ and $\displaystyle \frac{y}{\sqrt{x^2 + y^2}} = \sin\theta(x,y)$. So, it is the angle between the x-axis and the line connecting origo and the point $\displaystyle (x,y)$.

How do I show that the mapping $\displaystyle \theta(x,y)$ is not continous on the negative x-axis?

2. This mapping is discontinuous at (0,0).
Therefore, since you include this point in your mapping, it is discontinuous.
Remark that if your mapping would be $\displaystyle \theta : \mathbb{R}^2 \setminus \{ (x,y) | y = 0, x < 0 \} \rightarrow (-\pi,\pi)$ then it would be continuous.
Now to prove that the function is discontinuous at (0,0), you need to show that $\displaystyle \lim_{(x,y)\rightarrow (0,0)} \frac{x}{\sqrt{x^2+y^2}} \neq f(0,0)$
We don't care wath function f is since we can prove that the limit doesn't exist.
Let approach the point along the line of equation $\displaystyle y = mx$ The limit then become $\displaystyle \lim_{(x,y)\rightarrow (0,0)} \frac{x}{x\sqrt{1+m^2}} =\lim_{(x,y)\rightarrow (0,0)} \frac{1}{\sqrt{1+m^2}}$ Since the limit depends on m, by uniqueness of limit we conclude that the limit doesn't exist.
Finally, $\displaystyle \theta(x,y) = arccos(\frac{x}{\sqrt{x^2+y^2}})$ is not continuous since $\displaystyle \frac{x}{\sqrt{x^2+y^2}}$ is not.

3. ## No

I'm sorry, but it was not the point (0,0) I wanted to show a discontinouity at - this point is not part of the domain of the function. The domain was $\displaystyle \mathbb{R}^2\setminus\{ (x,y)| y=0,x \leq 0 \}$ - that is the whole x-y plane minus the negative x-axis, also minus the point (0,0).

I want to show that $\displaystyle \theta(x,y)$ is not continous on the negative x-axis (not including the point (0,0)). To prove that it is not possible to extend the function to a continous function on the domain $\displaystyle \mathbb{R}^2\setminus\{ (0,0) \}$.

It is easily seen that $\displaystyle \theta$ must obey that

$\displaystyle \lim_{y\rightarrow0^+} \theta(-1,y) \rightarrow \pi \qquad\text{and}\qquad \lim_{y\rightarrow0^-} \theta(-1,y) \rightarrow -\pi .$

for $\displaystyle x < 0$. But how do I show it?

You can see the system here: http://ragelse.com/file.php/8500c2bd...b9f3fc17f5d2da

4. Doesn't matter any more - I didn't had more time to do the assignment...