Let $f(x,y)=\frac{y^2-5x+x^2}{y-2x}$. Find the limit as $(x,y)$ approaches $(1,2)$.

$f(1,2)=\frac{0}{0}$, an indeterminate form. Therefore, direct substitution will not work.

By going along the path $x=1$, we have:

$\displaystyle\lim_{(x,y)\rightarrow(1,2)}\left(\frac{y^2-5x+x^2}{y-2x}\right)=\lim_{(x,y)\rightarrow(1,2)}\left(\frac{y^2-4}{y-2}\right) =\lim_{(x,y)\rightarrow(1,2)}(y+2)=4$.

Then, along the path $y=2$, we have:

$\displaystyle\lim_{(x,y)\rightarrow(1,2)}\left(\frac{y^2-5x+x^2}{y-2x}\right)=\lim_{(x,y)\rightarrow(1,2)}\left(\frac{x^2-5x+4}{-2x+2}\right) =\lim_{(x,y)\rightarrow(1,2)}(\frac{(x-4)(x-1)}{-2(x-1)})=\lim_{(x,y)\rightarrow(1,2)}\frac{(4-x)}{2}=\frac{3}{2}$.

Since the limit of $f(x,y)$ approaches two distinct values from different paths, so by the two path rule, $\displaystyle\lim_{(x,y)\rightarrow(1,2)}\left(\frac{y^2-5x+x^2}{y-2x}\right) \text{DNE}$.