# Multivariable Limit

1. Jul 9, 2013

### Bashyboy

"As $(x,y) \rightarrow (0,0)$, $r \rightarrow$," is a fact that I am given, in order to solve a problem. I simply want to know if I properly understand why this fact is true.

I know that $x = rcos \theta$ and $y=rsin \theta$. If I were to look at the individual limits, as $x \rightarrow 0$ and $y \rightarrow 0$, then I see each relation becomes $0 = rcos \theta$ and $0=rsin \theta$. Now, as x and y approach 0, there is no angle that $\theta$ approaches that make $rcos \theta$ and $rsin \theta$ simultaneously zero; hence, it must be that r approaches 0 as to make each expression zero simultaneously.

In summary, $\lim_{(x,y) \rightarrow (0,0)}~(x+y) = lim_{r \rightarrow 0} ~ (rcos \theta + rsin \theta)$.

Are my arguments correct?

2. Jul 9, 2013

### LCKurtz

You mean $r\rightarrow 0$, right?

What does $x+y$ have to do with anything? You are making something very simple seem very complicated. When you say $(x,y)\rightarrow (0,0)$ that means the distance from $(x,y)$ to $(0,0)$ goes to zero. That is $r$. Of course, maybe what you posted and what you were actually given aren't the same.

3. Jul 9, 2013

### Bashyboy

From my understanding, $(x,y) \rightarrow (0,0)$ means that the point (x,y) assumes the geometric location (0,0), not that the distance between two points becomes zero.

Also, x + y is just the addition of the two equations $x = r\cos \theta$ and $y=rsin \theta$

Last edited: Jul 9, 2013
4. Jul 9, 2013

### Mandelbroth

If we want to consider it as a limit here, we use distance. It can't just "assume" the location. We can make it arbitrarily close, though. :tongue:

Once you recognize that the limit is implying that the distance is going to zero, it becomes clear why your equation is correct, because r is the distance.

5. Jul 9, 2013

### Bashyboy

So, the distance becomes zero because the point (x,y) moves, as you said, arbitrarily close to (0,0)?

6. Jul 9, 2013

### Mandelbroth

No. This is a limit! The distance is allowed to become arbitrarily close to zero, thus making (x,y) arbitrarily close to (0,0).

7. Jul 9, 2013

### Bashyboy

I do not see any distinguishable qualities between what you just said and what I said in post #5. In short, you said precisely the same thing I did.

8. Jul 9, 2013

### Mandelbroth

The difference is your statement implies that $(x,y)=(0,0)$ and r=0, where r is the distance of the point (x,y) from (0,0). My statement implies that $\displaystyle\lim_{r\rightarrow 0}(x,y)=(0,0)$.

Last edited: Jul 9, 2013
9. Jul 9, 2013

### Bashyboy

Yes, but isn't it sometimes the case that the limit of a function at a particular value is equal to what the function is evaluated at that approaching value, thereby implying that the coordinate (x,y) has not just approached the limit, but is directly on said limit? This is decidedly true of polynomial functions.

10. Jul 9, 2013

### shortydeb

"The distance" never actually becomes zero. Are you familiar with the epsilon-delta definition of the limit? To truly understand the concept of a limit, you need to understand that definition.

11. Jul 10, 2013

### Bashyboy

Yes, I am acquainted with the epsilon-delta definition of a limit.

12. Jul 10, 2013

### HallsofIvy

The point is that there is a difference between being "arbitrarily close to (0, 0)" and "being at (0, 0)".

13. Jul 10, 2013

### LCKurtz

I am still wondering if you posted what you meant to post originally. Were you given $(x,y) \rightarrow (0,0)$ as you have written, or were you given that $x\rightarrow 0$ and $y\rightarrow 0$? They aren't the same statement and your question has more content if you were actually given the latter.

14. Jul 10, 2013

### Bashyboy

I was given that $(x,y) \rightarrow (0,0)$ implies that $r \rightarrow 0$

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