# Limit of f(x,y)

1. Mar 24, 2016

### says

1. The problem statement, all variables and given/known data
When we are taking a limit of a multivariable function, why do we use the points (x,0) and (0,y) to find out if the limit doesn't exist? They are two different points, no? If they are two different points then wouldn't they go to different points on the z-axis?

If we have a function, f(x,y), the output will be on the z-axis.
There are infinitely many ways in which we can get to a point on the surface (z-axis)

So if we start at the point (x,0) and apply the function we should end up at a point z.
and if we start at the point (0,y) and apply the function we should end up at the same point z.

Is that statement correct? It doesn't sound right. I'm having a little conceptual problem here so i'm trying to understand it in very simplistic language before i put it into more mathematically rigorous terms.

If we want to get to point C (f(x,y)), we can get there from point A (0,y) or point B (x,0). We would have to take different paths to get to C, but we WOULD get to C. If we found the C was different for A and B then we would say C (the limit) does not exist.

But then if there are infinitely many paths I can take to get to C, then doesn't that mean there are infinitely many points on the z-axis that I could go to? If that is so, then how do we know (0,y) and (x,0) should end up at the same point?

Sorry, I hope I haven't confused anyone with my question.

2. Relevant equations

3. The attempt at a solution

2. Mar 24, 2016

### LCKurtz

Most textbook problems on this topic give "easy" problems by asking for a limit as $(x,y)\to (0,0)$. I'm guessing that is the type of limit you are working with. If so, when you let x or y in your above example go to zero, the limiting point is $(x,y)=(0,0)$. You are just approaching it along two different paths. The point of trying that is that, in the case the limit doesn't exist, you might get lucky and get two different answers along those two paths. Then you would know the limit doesn't exist and you are done with the problem.

3. Mar 25, 2016

### says

But (x,0) and (0,y) are different points aren't they? so wouldn;t they point to different points of the surface?

4. Mar 25, 2016

### HallsofIvy

Staff Emeritus
"(x, 0)" and "(0, y)" are not points!. Since x and y are variables, they are lines- specifically the x and y axes. And you are taking the limits as x and y go to 0 so both going to (0, 0).