# Stuck Proving a Limit Doesn't exist

1. Feb 18, 2010

### rhololkeolke

1. The problem statement, all variables and given/known data
The problem is to either find the limit or show that it does not exist $$lim_{(x,y)\rightarrow(2,-2)}\frac{4-xy}{4+xy}$$

I've been able to do similar problems to this such as
$$lim_{(x,y)\rightarrow(0,0)}\frac{xy}{x^2+y^2}$$ where I took two different paths to the limit and found that they were not equal and so it didn't exist. However, for this one I can't seem to pick a function that gives me a limit that exists let alone two functions that give me two different limits.

I've tried coming from the following paths for the problem
$$y=-x$$
$$y=x-4$$
$$y=-2$$
$$y=x^2-6$$
$$x=2$$

No matter which one I do I can't seem to get anything to cancel out in order to simplify it to one that I can perform the limit on. Are there two functions that I can use to get this limit and how would I find these. If there aren't then how would I prove that this limit exists?

2. Feb 18, 2010

### Dick

I don't see anything wrong with y=(-x). Put that in and let x->2. What do you conclude?

3. Feb 18, 2010

### rhololkeolke

When I plug in y=-x the limit becomes
$$lim_{(x,y)\rightarrow(2,-2)}\frac{4+x^2}{4-x^2}$$
But then when I substitute in the values I still end up with 8/0 which is undefined. Am I doing something wrong when I plug in that function?

4. Feb 18, 2010

### Dick

You don't really just 'plug in'. You think about what happens as x approaches 2. And you are right, the numerator goes to 8 and the denominator goes to 0. I would say the limit doesn't exist. The quotient just gets larger and larger as x->2.

5. Feb 18, 2010

### rhololkeolke

So I don't have to show that multiple paths equal different things in this case just that one of the paths doesn't exist?

6. Feb 18, 2010

### Dick

Right. In this case there is no limit along ANY path. No matter what the path you still get an 8/0 type quotient.

7. Feb 18, 2010

### rhololkeolke

Thank you. That makes so much sense now. I was under the impression that we had to find two functions with different values but now I can see that this works too.