Problem 22
Question
Evaluate the following limits. $$\lim _{(x, y) \rightarrow(-1,1)} \frac{2 x^{2}-x y-3 y^{2}}{x+y}$$
Step-by-Step Solution
Verified Answer
Question: Determine if the given limit exists and find its value, if it does:
$$\lim_{(x, y) \to (-1, 1)} \frac{2x^2 - xy - 3y^2}{x + y}$$
Answer: The given limit does not exist.
1Step 1: Factor the numerator
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:
$$\frac{(2x + y)(x - 3y)}{x + y}$$
2Step 2: Check the denominator
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.
3Step 3: Check if the limit is path-independent
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.
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