Problem 21 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(3,1)} \frac{x^{2}-7 x y+12 y^{2}}{x-3 y}$$

Step-by-Step Solution

Answer

Answer: The limit is -1.

Step 1: Factor the numerator

In order to further simplify the function, we need to factor the numerator of the given expression: $$\lim _{(x, y) \rightarrow(3,1)} \frac{x^{2}-7 x y+12 y^{2}}{x-3 y}$$ The numerator has the form of a quadratic expression, so we will look for two numbers that multiply to 12y^2 and add up to -7xy. We can factor the numerator as follows: $$ x^2 - 7xy + 12y^2 = (x - 3y)(x - 4y) $$ So, the expression becomes: $$\lim _{(x, y) \rightarrow(3,1)} \frac{(x - 3y)(x - 4y)}{x-3 y}$$

Step 2: Simplify the expression

Now that we have factored the numerator, we can simplify the fraction by canceling out the common factor with the denominator: $$\lim _{(x, y) \rightarrow(3,1)} \frac{(x - 3y)(x - 4y)}{x-3 y} = \lim _{(x, y) \rightarrow(3,1)} (x - 4y)$$

Step 3: Substitute the values of x and y to find the limit

The expression has been simplified, and we can now directly substitute the values x = 3 and y = 1 into the expression: $$\lim _{(x, y) \rightarrow(3,1)} (x - 4y) = (3 - 4 \cdot 1) = 3 - 4 = -1$$ Now that we have substituted the values of x and y, we have found the limit: $$\lim _{(x, y) \rightarrow(3,1)} \frac{x^{2}-7 x y+12 y^{2}}{x-3 y} = -1$$