Prove that the limit of the following function of two variables is zero I need to prove the following:

$$\lim_{(x,y)\to (1 ,2)} \frac{x^2+2xy-6x-2y+5}{\sqrt{(x-1)^2+(y-2)^2}}=0$$

I've tried to solve it by substituting $y=mx$ but I can't get the solution that way. Any help would be appreciated thanks.
 A: Hint
You can use $\varepsilon-\delta$ definition and find that limit, but I'd just switch to polar coordinates
\begin{align}
x = 1 + r \cos \phi \\
y = 2 + r \sin \phi
\end{align}
so the whole expression becomes 
\begin{align}
\lim_{(x,y) \to (1,2)} (\ldots) = \lim_{r \to 0^+} r^2 (\cos \phi + 2 \sin \phi)
\end{align}
The latter to exist you should prove that it doesn't depend on $\phi$, which shouldn't be hard due to the boundedness of $\sin$ and $\cos$ functions.
A: Factor the numerator as $$x^2+2xy-6x-2y+5=(x-1)[(x-1)+2(y-2)]$$Then, use the triangle inequality$$|(x-1)[(x-1)+2(y-2)]|\le|(x-1)^2|+2|x-1||y-2|$$to show that$$\frac{|x^2+2xy-6x-2y+5|}{\sqrt{(x-1)^2+(y-2)^2}}\le\frac{|(x-1)^2|+2|x-1||y-2|}{\sqrt{(x-1)^2+(y-2)^2}}$$Noting that $\sqrt{(x-1)^2+(y-2)^2}\ge|x-1|$ $$\frac{|x^2+2xy-6x-2y+5|}{\sqrt{(x-1)^2+(y-2)^2}}\le\frac{|(x-1)^2|+2|x-1||y-2|}{\sqrt{(x-1)^2+(y-2)^2}}\le|x-1|+2|y-2|$$
Finally, for all $\epsilon>0$ $$\frac{|x^2+2xy-6x-2y+5|}{\sqrt{(x-1)^2+(y-2)^2}}<\epsilon$$whenever $|x-1|<\delta_1=\frac{\epsilon}{3}$ and $|y-2|<\delta_2=\frac{\epsilon}{3}$.
