$\frac{|x|-|y|}{|x-y|}$ 
Let $x\neq y$. What values $\frac{|x|-|y|}{|x-y|}$ can take?

I tried: If $|x|>|y|$, it is $1$, if $|x|<|y|$, it is $-1$ and if $|x|=|y|$, it is $0$. Have I missed anything?
 A: We have $|x-y| \geq ||x| - |y||$ so $$ \frac{|x|-|y|}{|x-y|} \in [-1,1] $$ for all $x,y \in \Bbb R$, $x \neq y$. 
Moreover, every element $a \in [-1,1]$ can be obtained. You already found $1$, $-1$ and $0$.
For $a \in (0,1)$, setting $x = -\frac{1}{2}(\frac{1}{a} + 1)$ and $y = \frac{1}{2}(\frac{1}{a} - 1)$ gives $$|x-y| = \frac{1}{a}$$ and $$|x| - |y| = 1.$$
For $a \in (-1,0)$ you can take $x = \frac{1}{2}(\frac{1}{|a|} - 1)$ and $y = -\frac{1}{2}(\frac{1}{|a|} + 1)$ to obtain $$ |x-y| = \frac{1}{|a|} $$ and $$|x| - |y| = -1.$$
In any case, you get $\frac{|x|-|y|}{|x-y|} = a$ as desired.
A: Let $y=0$ and $x=1$.
Thus, we get a value $1$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\frac{|x|-|y|}{|x-y|}\leq1$$ or
$$|x|-|y|\leq|x-y|$$ or
$$|x-y|+|y|\geq|x|,$$
which is true by the triangle inequality:
$$|x-y|+|y|\geq|x-y+y|=|x|.$$
For $x=0$ and $y=-1$ we get a value $-1$.
We'll prove that it'a minimal value.
Indeed, we need to prove that
$$\frac{|x|-|y|}{|x-y|}\geq-1$$ or
$$|x|-|y|\geq-|x-y|$$ or
$$|x-y|+|x|\geq|y|,$$ which is true by the triangle inequality again:
$$|x-y|+|x|=|x-y|+|-x|\geq|x-y-x|=|y|.$$
Id est, $$-1\leq \frac{|x|-|y|}{|x-y|}\leq1.$$
Now, let $x>0$, $k\geq0$ and $y=-kx$.
Thus, $$\frac{|x|-|y|}{|x-y|}=\frac{1-k}{1+k}$$
and since for all $l\in(-1,1]$ there is $k\geq0$ for which $-1<\frac{1-k}{1+k}\leq1$, 
we obtain that $\frac{|x|-|y|}{|x-y|}$ can get all value from $[-1,1]$.
