Show that $\lim\limits_{r\to\infty} \textrm{\{} \|x-ru\|-\|y-ru\| \textrm{}\} = \left$ 
Let $x,y,u \in \mathbb{R}^2, r\in\mathbb{R}$ and $\|\cdot\|$ be the norm. Show that $$\lim\limits_{r\to\infty} \textrm{\{} \|x-ru\|-\|y-ru\| \textrm{}\} = \left<y-x,u\right>$$

I have to tried to use the polarization identity but am still stuck. I am not able to get rid of the variable $r$ or remove the limit. Any help or insight is appreciated. Maybe the RHS of the above is not in the correct/accurate form, as in my analysis $||u||=1$, but more or less RHS should be achieved in a similar form.
 A: Putting $\varepsilon=\frac{1}{|r|}$ and $\delta=-2\frac{\left< x,u \right>}{\|u\|^2}\varepsilon +\varepsilon^2$, one has
$$
\begin{array}{lcl}
\| x-ru \| &=& \sqrt{ \| x-ru \|^2 } \\
 &=& \sqrt{ \| x \|^2 -2r \left< x,u \right> +r^2 \|u\|^2  } \\
 &=& |r| \| u \| \sqrt{ 1 -2\frac{\left< x,u \right>}{\|u\|^2}\varepsilon +\varepsilon^2  } \\
  &=& |r| \| u \| \sqrt{ 1 +\delta } \\
 &=& |r| \| u \| \bigg( 1 +\frac{\delta}{2}+o(\delta)  \bigg) \\
 &=& |r| \| u \| \bigg( 1 -\frac{\left< x,u \right>}{\|u\|^2}\varepsilon+o(\varepsilon) \bigg) \\
  &=& |r| \| u \|  -\frac{\left< x,u \right>}{\|u\|}+o(1)  
  \\
\end{array}
$$
Substracting, one deduces 
$$
\| x-ru \|-\| y-ru \|=\frac{\left< y-x,u \right>}{\|u\|}+o(1).  
$$
A: Here $||u||=1$. Then $\lim\limits_{r\to\infty} \textrm{\{} \|x-ru\|-\|y-ru\| \textrm{}\} = \lim\limits_{r\to\infty} \textrm{\{} \frac{r[||x/r||^2-||y/r||^2+2<((y-x)/r,u>|]}{||x/r-u||+||y/r-u||} \textrm{}\}=<y-x,u>$. I have made use of the polarization identity. 
A: \begin{align}
   \|x-ru\|-\|y-ru\| \textrm{}
   &= \sqrt
   {\langle x,x \rangle - 2r\langle x,u \rangle + r^2\langle u,u \rangle}
   -\sqrt
   {\langle y,y \rangle - 2r\langle y,u \rangle + r^2\langle u,u \rangle }
\\
   &= \frac
   {(\langle x,x \rangle - 2r\langle x,u \rangle + r^2\langle u,u \rangle)
   -(\langle y,y \rangle - 2r\langle y,u \rangle + r^2\langle u,u \rangle)}
   {\sqrt{\langle x,x \rangle - 2r\langle x,u \rangle + r^2\langle u,u \rangle }
   +\sqrt{\langle y,y \rangle - 2r\langle y,u \rangle + r^2\langle u,u \rangle }}\\
   &= \frac
   {\langle x,x \rangle - \langle y,y \rangle  + 2r\langle y-x,u \rangle}
   {\sqrt{\langle x,x \rangle - 2r\langle x,u \rangle + r^2\langle u,u \rangle }
   +\sqrt{\langle y,y \rangle - 2r\langle x,y \rangle + r^2\langle u,u \rangle }}
\\
   &= \frac
   {
      \dfrac
         {\langle x,x \rangle- \langle y,y \rangle}
         {r} 
      + 2\langle y-x,u \rangle
   }
   {
      \sqrt
         {
            \dfrac{\langle x,x \rangle}{r^2}
            - 2\dfrac{\langle x,u \rangle}{r}
            + \langle u,u \rangle
         }
   +\sqrt
      {
         \dfrac{\langle y,y \rangle}{r^2}
         - 2\dfrac{\langle x,y \rangle}{r}
         + \langle u,u \rangle
      }
   }
\\
   &\to \frac{\langle y-x,u \rangle}{\|u\|} \text{ as } r \to \infty
\end{align}
