Stuck Proving a Limit Doesn't exist

[rhololkeolke]

Homework Statement

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?

 

[Dick]

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

 

[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?

 

[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.

 

[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?

 

[Dick]

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?

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

 

[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.