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Chapter 12: Problem 24

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

Short Answer

Expert verified
Answer: The limit of the function as \((x, y) \rightarrow (4,5)\) is \(\frac{1}{6}\).

Step by step solution

01

Analyze the given function

Consider the given function: $$ f(x, y) = \frac{\sqrt{x+y}-3}{x+y-9} $$ First, notice that if we attempt to find the limit as \((x, y) \rightarrow (4,5)\) without any simplification, we will encounter an indeterminate form because the numerator and denominator will both be zero. Therefore, we first need to manipulate the function algebraically to remove the indeterminate form.
02

Manipulate the function

Let's multiply both numerator and denominator by the conjugate of the numerator: $$ \frac{\sqrt{x+y}-3}{x+y-9} \times \frac{\sqrt{x+y}+3}{\sqrt{x+y}+3} = \frac{(x+y)-9}{(x+y-9)(\sqrt{x+y}+3)} $$ We can now simplify the function as follows: $$\frac{(x+y)-9}{(x+y-9)(\sqrt{x+y}+3)}$$ Now, as \((x, y) \rightarrow (4,5)\), the factor \((x+y-9)\) would become 0. To remove this indeterminate form, we can cancel out this factor from the numerator and the denominator.
03

Cancel the common factor and find the limit

We cancel the factor of \((x+y-9)\) from both the numerator and the denominator, resulting in the simplified function: $$g(x, y) = \frac{1}{\sqrt{x+y}+3}$$ Now that we have removed the indeterminate form, we can directly substitute the values \((x, y) \rightarrow (4,5)\) into the simplified function to find the limit: $$\lim_{(x, y) \rightarrow(4,5)} g(x, y) = \frac{1}{\sqrt{4+5}+3}$$
04

Evaluate the limit

Now, we can evaluate the limit by plugging in the values: $$\frac{1}{\sqrt{9}+3} = \frac{1}{3+3} = \frac{1}{6}$$ Thus, the limit of the function as \((x, y) \rightarrow (4,5)\) is \(\frac{1}{6}\).

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