Use the two$-$path lest to prove that the following limit does not exist.

$\underset{(xy)(0,0{)}^x}{lim}\frac{x2y}{x-2y}$

A$.$ Exy$)$ approaches $1.($Simplify your answer.$)$

B$.$ fixy$)$ has no limit and does not approach $$ or $-$ as $($x$,$y$)$ approaches $(0,0)$ along the $x-$axis.

What value does $f(x,y)=\frac{x2y}{x-2y}$ approach as $($x$.$y$)$ approaches $(0,0)$ aftong the y$-$axis? Select the correct choice below and, II necessary, fill in the answer box to complete your choice.

A$.f(xy)$ approaches $($Simplify your answer$)$

B$.$ fixy$)$ has no limit and does not approach $$ or $-$ as $($x$.$y$)$ approaches $(0,0)$ along the y$-$anh.

Why does the given limit not exist?

A$.$ As $(x,y)$ approaches $(0,0)$ along different paths, fix.y$)$ does not always approach a finite value.

$8.$ The limit does not exist because as $($x$,$y$)$ approaches $(0,0),$ the denominator approaches $0.$

C$.$ As $($x$.$y$)$ approaches $(0\mathrm{.}0)$ along different paths, fixy$)$ approaches two different values

D$.$ As $($x$,$y$)$ approaches $(0\mathrm{.}0)$ along different paths, fif, y$)$ always approaches the same yatur