Lim x^2-4y^2/(x+2y) as (x,y)->(-2,1): -4 or DNE?

In summary, the limit of the function as (x,y) approaches (-2,1) is -4. Yes, the limit is defined and equals -4. To find the limit, substitute the given values of x and y into the expression and simplify. In this case, we get (-2)^2-4(1)^2/(-2+2(1)) = -4/0, which simplifies to -4. The limit equals -4 because as (x,y) approaches (-2,1), the denominator of the expression (x+2y) becomes smaller and smaller, and the numerator (x^2-4y^2) remains constant. This results in a value of -4, which
  • #1
Dustinsfl
2,281
5
lim (x^2-4y^2)/(x+2y) as (x,y)->(-2,1)
Does this limit go to -4 or DNE since it is undefined all along x+2y?
 
Physics news on Phys.org
  • #2
It Does Not Exist.
 
  • #3
Please note that you may write this as:
[tex]\frac{x^{2}-4y^{2}}{x+2y}=\frac{(x-2y)(x+2y)}{x+2y}=x-2y,x\neq{-}2y[/tex]
 

Related to Lim x^2-4y^2/(x+2y) as (x,y)->(-2,1): -4 or DNE?

1. What is the limit of the function (x,y) → (-2,1) of the expression x^2-4y^2/(x+2y)?

The limit of the function as (x,y) approaches (-2,1) is -4.

2. Is the limit of the function (x,y) → (-2,1) of the expression x^2-4y^2/(x+2y) defined?

Yes, the limit is defined and equals -4.

3. How do you find the limit of the function (x,y) → (-2,1) of the expression x^2-4y^2/(x+2y)?

To find the limit, substitute the given values of x and y into the expression and simplify. In this case, we get (-2)^2-4(1)^2/(-2+2(1)) = -4/0, which simplifies to -4.

4. Why does the limit of the function (x,y) → (-2,1) of the expression x^2-4y^2/(x+2y) equal -4?

The limit equals -4 because as (x,y) approaches (-2,1), the denominator of the expression (x+2y) becomes smaller and smaller, and the numerator (x^2-4y^2) remains constant. This results in a value of -4, which is the limit.

5. Can the limit of the function (x,y) → (-2,1) of the expression x^2-4y^2/(x+2y) be calculated in any other way?

No, the limit can only be calculated by substituting the given values of x and y into the expression and simplifying. This is because the limit represents the value that the function approaches as the input values approach the given point, and there is no other general method for calculating this value.

Similar threads

Replies
3
Views
3K
Replies
1
Views
241
Replies
6
Views
2K
  • Calculus
Replies
9
Views
1K
Replies
1
Views
970
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
Replies
8
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
669
Back
Top