How to solve multivariable limits using various methods?

In summary, the student attempted to find a solution to the homework equation by substituting different values, factorizing the denominator, and simplifying the equation.
  • #1
RJLiberator
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Homework Statement



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

Homework Equations



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

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?
 
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  • #2
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?
 
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  • #3
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
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?
 
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  • #5
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 :eek:.
 
  • #6
You're on the right path.
 
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  • #7
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
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.
 
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  • #9
Indeed, excellent rigorous observation.
 

Related to How to solve multivariable limits using various methods?

What are multivariable limits?

Multivariable limits are limits of functions with more than one independent variable. In other words, they involve finding the value that a function approaches as multiple variables approach a certain point.

Why are multivariable limits important?

Multivariable limits are important because they allow us to understand the behavior of a function in multiple dimensions. They help us determine if a function is continuous, differentiable, and has a maximum or minimum value at a certain point.

How do I find multivariable limits?

To find multivariable limits, you need to evaluate the function at the given point and see if the value exists and is the same regardless of the path taken to approach that point. If the values are different, the limit does not exist.

What is the difference between a multivariable limit and a single variable limit?

The main difference between a multivariable limit and a single variable limit is the number of variables involved. Single variable limits only consider the behavior of a function as one variable approaches a point, while multivariable limits consider the behavior as multiple variables approach a point simultaneously.

Can multivariable limits have different values depending on the path taken?

Yes, multivariable limits can have different values depending on the path taken. This is because the behavior of a function can vary in multiple dimensions, so the limit may exist along one path but not along another. This is why it is important to consider all possible paths when evaluating a multivariable limit.

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