# Help me with multivariable limits

1. Feb 13, 2015

### RJLiberator

1. The problem statement, all variables and given/known data

Lim (x,y,)--> (1, 3) of (x^2-1)/(xy-y)

2. Relevant equations

I know that the answer is 2/3 according to wolfram alpha multivariable limit calculator.

3. The attempt at a solution

So this is my first time doing multivariable limits, I've studied the following:

1) Direct substitution : Well, this doesn't work in our case as we get 0/0.
2) Try to go along different paths: say y=x and so forth. This also doesn't work. If we make y=x then we get 0/0. And I also don't feel good about trying to prove that the limit exists with this path idea.
3) The 'squeeze' idea: I'm not really sure how to apply this to this particular problem. Is this what I need to do?

Is there anything else that I am not seeing?

Last edited: Feb 13, 2015
2. Feb 13, 2015

### wabbit

One way to get past a 0/0 answer is to find if it can be rewritten as 0*a/0*b,and then see if the 0 part can be taken out. Have you tried that?

3. Feb 13, 2015

### RJLiberator

No, I have not.

So 0*a/0*b, hm.
Do you mean such as setting x=0 where it becomes
(0)^2-1/((0)*y)-y)
And then we get the answer of -1/-3 which is 1/3?

This seems to work out well! However, is this enough to prove that the limit does indeed exist?

4. Feb 13, 2015

### wabbit

Hmm no, not quite, why set x=0?
No, by "0" here I mean for instance " the value of xy - y at (x=1,y=3). Could that turn out to be a product?

5. Feb 13, 2015

### RJLiberator

I was setting x=0 to go along that path? :/

xy-y = 1(3)-(3) = 0 ?
y(x-1) = 3(1-1) = 0?

I'm not sure what you mean by turn out to be a product .

6. Feb 13, 2015

### wabbit

You're on the right path.

7. Feb 13, 2015

### RJLiberator

AHHHHHH. I was thinking way too much.
Factor the denominator and numerator.
Simplifies to
(x+1)/3
plug in values, and walouh, 2/3 is the answer I was looking for.

BEAUTIFUL. Thank you for your guidance.

8. Feb 13, 2015

### wabbit

Right. It's a little better if you factor to (x+1)/y and only substitute at the end because then you can see the result is true whichever way (x,y) goes to (1,3), while you only really proved it if y goes to 3 first, then x goes to 1. But you've got it now.

9. Feb 13, 2015

### RJLiberator

Indeed, excellent rigorous observation.