91Ó°ÊÓ

First, we factor the numerator to possibly simplify the expression. The given expression is: $$\frac{2 x^{2}-x y-3 y^{2}}{x+y}$$ Factor the numerator: $$\frac{(2x + y)(x - 3y)}{x + y}$$

Now, let's check if the denominator approaches zero as (x, y) approaches (-1, 1). Plug in the values into the denominator: $$x + y = -1 + 1 = 0$$ Since the denominator approaches zero, we cannot directly divide the numerator by the denominator and substitute the values of x and y to find the limit.

To check if the limit is path-independent, let's approach the point (-1, 1) through different paths and check if the limit is the same: Path 1: y = 1 - x Replace y by (1 - x) in the given expression and find the limit as x approaches -1. $$\frac{(2x + (1 - x))(x - 3(1 - x))}{x + (1 - x)}$$ Cancel out the denominator and the term (x + 1 - x) from numerator and denominator: $$\frac{(-x)(x - 3(1 - x))}{1}$$ Now we can calculate the limit as x approaches -1: $$\lim_{x \to -1} \frac{-x(3x-2)}{1} = \frac{-1(-5)}{1} = 5$$ Path 2: x = -1 - y Replace x by (-1 - y) in the given expression and find the limit as y approaches 1. $$\frac{(2(-1 - y) + y)((-1 - y) - 3y)}{(-1 - y) + y}$$ Cancel out the denominator and the term (-1 - y + y) from numerator and denominator: $$\frac{(-2y-1)((-1 - y) - 3y)}{-1}$$ Now we can calculate the limit as y approaches 1: $$\lim_{y \to 1} \frac{-3(2y - 1)}{-1} = \frac{-3(1)}{-1} = 3$$ Since the limits through the two paths are different, the two-variable limit does not exist. Therefore, the given limit is: $$\lim_{(x, y) \to (-1, 1)} \frac{2x^2 - xy - 3y^2}{x + y}$$ does not exist.