Two paths that show that $\frac{x-y}{x^2 + y^2}$ has no limit when $(x,y) \rightarrow (0,0)$ I'm having a difficult time trying to find two different paths that give me different limits  for the following:
$$\lim_{(x,y) \rightarrow (0,0)}  \frac{x-y}{x^2 + y^2}$$
 A: For reasons that will become apparent further down, I find it useful to instead study $$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{x^2+y^2}{x-y}\right).$$
Note that for all $(x,y)\in \mathbb R^2$ with $x\neq y$ it holds that $$\dfrac{x^2+y^2}{x-y}=\dfrac{2x^2-(x^2-y^2)}{x-y}=\dfrac{2x^2}{x-y}-(x+y).$$
Since $\lim \limits_{(x,y)\to (0,0)}(x+y)$ exists, any problems that might occur with the initial limit will occur on $$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^2}{x-y}\right).$$
This last limit is fairly simple to manipulate by choosing $y$ as a function of $x$. If you stare at the fraction long enough, and hopefully not for too long, you'll see that setting $y=x-kx^2$ with $k\in \mathbb R$ will yield a $k$-dependent limit.
You can now abandon this scratch work and consider the paths $t\mapsto (t,t-kt^2)$ in the initial limit.
A: For some problems with such a great degree symmetry in $x$ and $y$ you can try working in polar coordinates
\begin{equation*}
\begin{cases}
x = r \cos(\theta) \\
y = r \sin(\theta)
\end{cases}
\end{equation*}
The statement $(x,y) \rightarrow (0,0)$ is equivalent to $r \rightarrow 0$, and by taking $\theta = \theta(r)$ you get any path you want.
Now, the limit in terms of polar coordinates is
\begin{equation*}
\lim_{r \rightarrow 0}\frac{\cos(\theta) - \sin(\theta)}{r}
\end{equation*}
If you pick a path with $\theta$ such that $\cos(\theta)\neq \sin(\theta)$, you get $\infty$.
If you pick a path with $\theta$ such that $\cos(\theta) = \sin(\theta)$, you get $0$.
