Problem 19 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(6,2)} \frac{x^{2}-3 x y}{x-3 y}$$

Step-by-Step Solution

Answer

Answer: The limit of the given function as \((x, y) \rightarrow (6, 2)\) is 6.

Step 1: Substitute the point (6, 2) into the given function

Given the function \(f(x, y) = \frac{x^{2}-3xy}{x-3y}\), we want to find $$\lim _{(x, y) \rightarrow(6,2)} \frac{x^{2}-3 x y}{x-3 y}$$. At first, we try to substitute (6, 2) into the function: $$f(6, 2) = \frac{6^2 - 3(6)(2)}{6 - 3(2)}$$

Step 2: Check for factorization

Before substituting the numbers, let's check if the function can be factorized: $$\frac{x^{2}-3 x y}{x-3 y} = \frac{x(x-3y)}{x-3y}$$ Now we see that there is a common factor of \((x-3y)\) in both the numerator and the denominator of the function.

Step 3: Cancel the common factors

Cancel the common factors of \((x-3y)\) and simplify the function: $$f(x, y) = x$$

Step 4: Substitute the point (6, 2) into the simplified function

Now, we can substitute the point (6, 2) into the simplified function: $$f(6, 2) = 6$$ Thus, the limit of the given function as \((x, y) \rightarrow (6, 2)\) is: $$\lim _{(x, y) \rightarrow(6,2)} \frac{x^{2}-3 x y}{x-3 y} = 6$$