Problem 23 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(2,2)} \frac{y^{2}-4}{x y-2 x}$$

Step-by-Step Solution

Answer

Answer: The limit is 2.

Step 1: Identify the limit point and rewrite the limit expression

Given the limit expression, we can see that the limit point is \((2, 2)\). Let's rewrite the expression as: $$\lim _{(x, y) \rightarrow(2,2)} \frac{y^{2}-4}{x y-2 x}$$

Step 2: Plug in the limit point into the function

Plug in the coordinates \((2, 2)\) into the function to identify the form: $$\frac{2^2-4}{(2)(2) - 2(2)} = \frac{4-4}{4-4} = \frac{0}{0}$$ Since we get an indeterminate expression of the form \(\frac{0}{0}\), we need to simplify the expression further.

Step 3: Simplify the numerator and denomitor before attempting to solve for the limit

Factor the numerator and the denominator: $$\frac{y^2 - 4}{xy - 2x} = \frac{(y + 2)(y - 2)}{x(y - 2)}$$

Step 4: Cancel out common factors

Divide the \((y-2)\) terms in the numerator and denominator to further simplify the expression: $$\frac{(y + 2)(y - 2)}{x(y - 2)} = \frac{y + 2}{x}$$

Step 5: Plug in the limit point again after simplification

Now, plug in the point \((2, 2)\) into the simplified expression: $$\lim _{(x, y) \rightarrow(2,2)} \frac{y + 2}{x} = \frac{2 + 2}{2} = \frac{4}{2}$$

Step 6: Evaluate the limit

Now that the expression has been simplified, the limit becomes a simple calculation: $$\frac{4}{2} = 2$$ So, the limit of the given function as \((x, y) \rightarrow (2, 2)\) is \(2\).