Problem 24 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(4,5)} \frac{\sqrt{x+y}-3}{x+y-9}$$

Step-by-Step Solution

Answer

Question: Evaluate the two-variable limit: $$\lim _{(x, y) \rightarrow(4,5)} \frac{\sqrt{x+y}-3}{x+y-9}$$ Answer: \(\frac{1}{6}\)

Step 1: Simplify the expression.

Our goal is to find a simplification of the expression that will allow us to find the limit as \((x,y) \rightarrow (4,5)\). Our first goal is to see if the denominator contains any factors that can be canceled out when simplified. To do that, we need to factor the denominator. However, in this case, the denominator, \(x+y-9\), does not factor so we move on to the next step.

Step 2: Rationalize the numerator.

In some problems, we can cancel out factors in the denominator by rationalizing the numerator. To do this, we can multiply the expression by the conjugate of the numerator. The conjugate of \(\sqrt{x+y}-3\) is \(\sqrt{x+y}+3\). Multiply both the numerator and denominator by the conjugate: $$\frac{\sqrt{x+y}-3}{x+y-9}\times \frac{\sqrt{x+y}+3}{\sqrt{x+y}+3}$$

Step 3: Multiply and simplify.

Now, let's simplify the expression by multiplying the numerators and the denominators: $$\frac{(\sqrt{x+y}-3)(\sqrt{x+y}+3)}{(x+y-9)(\sqrt{x+y}+3)}$$ Using the difference of squares formula \((a-b)(a+b) = a^2 - b^2\), we get: $$\frac{(\sqrt{x+y})^2 - 3^2}{(x+y-9)(\sqrt{x+y}+3)}$$ We simplify the numerator: $$\frac{x+y-9}{(x+y-9)(\sqrt{x+y}+3)}$$

Step 4: Cancel out common factors.

We can now cancel out the common factor of \(x+y-9\) in the numerator and denominator: $$\frac{x+y-9}{(x+y-9)(\sqrt{x+y}+3)} \times \frac{1}{x+y-9} = \frac{1}{\sqrt{x+y}+3}$$

Step 5: Evaluate the limit.

Now let's substitute the coordinates \((4,5)\) into the simplified expression to find the limit: $$\lim _{(x, y) \rightarrow(4,5)} \frac{1}{\sqrt{x+y}+3} = \frac{1}{\sqrt{4+5}+3}=\frac{1}{\sqrt{9}+3} = \frac{1}{3+3} = \frac{1}{6}$$ The given limit is evaluated as follows: $$\lim _{(x, y) \rightarrow(4,5)} \frac{\sqrt{x+y}-3}{x+y-9} = \frac{1}{6}$$