Evaluate the limit. \[ \lim _{(x, y) \rightarrow(1,3)} \frac{\sqrt{x+y}-2}{x+y-4} \] Find an expression that is equal to $f(x, y)=\frac{\sqrt{x+y}-2}{x+y-4}$, for all $(x, y)$ in the domain of $f$, that will be better suited to find the limit, if it exists. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. $\frac{\sqrt{x+y}-2}{x+y-4}=\square$ for all $(x, y)$ in the domain of $f$. (Simplify your answer.) B. The function $f(x, y)$ is already in a form suitable for finding the limit, if it exists. Evaluate the limit. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. $\lim _{(x, y) \rightarrow(1,3)} \frac{\sqrt{x+y}-2}{x+y-4}=\square$ (Simplify your answer.) B. The limit does not exist.

Step 1 ：First, we notice that the function \(f(x, y)=\frac{\sqrt{x+y}-2}{x+y-4}\) is undefined at the point (1,3). This is because the denominator of the fraction becomes zero at this point, which makes the function undefined.

Step 2 ：To find the limit as (x, y) approaches (1,3), we need to find an equivalent expression for \(f(x, y)\) that is defined at (1,3). We can do this by multiplying the numerator and denominator of the fraction by the conjugate of the numerator.

Step 3 ：\(f(x, y)=\frac{\sqrt{x+y}-2}{x+y-4} = \frac{(\sqrt{x+y}-2)(\sqrt{x+y}+2)}{(x+y-4)(\sqrt{x+y}+2)}\)

Step 4 ：Simplify the expression, we get \(f(x, y)=\frac{x+y-4}{(x+y-4)(\sqrt{x+y}+2)}\)

Step 5 ：Cancel out the common factor in the numerator and denominator, we get \(f(x, y)=\frac{1}{\sqrt{x+y}+2}\) for all (x, y) in the domain of f, except for (1,3).

Step 6 ：Now, we can find the limit as (x, y) approaches (1,3) using this new expression for f(x, y).

Step 7 ：\(\lim _{(x, y) \rightarrow(1,3)} \frac{1}{\sqrt{x+y}+2} = \frac{1}{\sqrt{1+3}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}\)

Step 8 ：So, \(\lim _{(x, y) \rightarrow(1,3)} \frac{\sqrt{x+y}-2}{x+y-4} = \boxed{\frac{1}{4}}\)