Use the two-path test to prove that the following limit does not exist.
\[
\lim _{(x, y) \rightarrow(0,0)} \frac{x+2 y}{x-2 y}
\]
What value does $f(x, y)=\frac{x+2 y}{x-2 y}$ approach as $(x, y)$ approaches $(0,0)$ along the $x$-axis? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. $f(x, y)$ approaches $\square$.
(Simplify your answer.)
B. $f(x, y)$ has no limit and does not approach $\infty$ or $-\infty$ as $(x, y)$ approaches $(0,0)$ along the $x$-axis.
What value does $f(x, y)=\frac{x+2 y}{x-2 y}$ approach as $(x, y)$ approaches $(0,0)$ along the $y$-axis? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. $f(x, y)$ approaches
(Simplify your answer.)
B. $f(x, y)$ has no limit and does not approach $\infty$ or $-\infty$ as $(x, y)$ approaches $(0,0)$ along the $y$-axis.
Why does the given limit not exist?
A. The limit does not exist because as $(x, y)$ approaches $(0,0)$, the denominator approaches 0 .
B. As $(x, y)$ approaches $(0,0)$ along different paths, $f(x, y)$ approaches two different values.
C. As $(x, y)$ approaches $(0,0)$ along different paths, $f(x, y)$ always approaches the same value.
D. As $(x, y)$ approaches $(0,0)$ along different paths, $f(x, y)$ does not always approach a finite value.

Step 1 ：Consider the function \(f(x, y)=\frac{x+2 y}{x-2 y}\).

Step 2 ：We first find the limit of the function as \((x, y)\) approaches \((0,0)\) along the x-axis. This means we set \(y=0\) in the function.

Step 3 ：Substituting \(y=0\) into the function, we get \(f(x, 0)=\frac{x+2(0)}{x-2(0)}=\frac{x}{x}=1\).

Step 4 ：So, the limit of the function as \((x, y)\) approaches \((0,0)\) along the x-axis is \(\boxed{1}\).

Step 5 ：Next, we find the limit of the function as \((x, y)\) approaches \((0,0)\) along the y-axis. This means we set \(x=0\) in the function.

Step 6 ：Substituting \(x=0\) into the function, we get \(f(0, y)=\frac{0+2y}{0-2y}=\frac{2y}{-2y}=-1\).

Step 7 ：So, the limit of the function as \((x, y)\) approaches \((0,0)\) along the y-axis is \(\boxed{-1}\).

Step 8 ：Since the limits along the x-axis and y-axis are different, the limit of the function as \((x, y)\) approaches \((0,0)\) does not exist.

Step 9 ：This is because the definition of the limit requires the function to approach the same value regardless of the path taken to \((0,0)\).

Step 10 ：So, the limit does not exist because as \((x, y)\) approaches \((0,0)\) along different paths, \(f(x, y)\) approaches two different values.

Step 11 ：Final Answer: The limit of the function as \((x, y)\) approaches \((0,0)\) along the x-axis is \(\boxed{1}\). The limit of the function as \((x, y)\) approaches \((0,0)\) along the y-axis is \(\boxed{-1}\). The limit does not exist because as \((x, y)\) approaches \((0,0)\) along different paths, \(f(x, y)\) approaches two different values.