Q:

Limit of $x^2-4xy+y^2$ as $(x, y)$ go to infinity

Can someone please tell me what the limit is of the following function as $x,y$ go to infinity?
$$x^2-4xy+y^2.$$
I think it is $-\infty$, but not sure. thanks

A:

Hint
Consider the case $x=y$ and the case $x=y^2$.

A:

The limit does not exist. If we choose $y=x$, with $x$ large positive, then $x^2-4xy+y^2$ is large negative. But if we choose $y=(2\pm\sqrt{3})x$, and $x$ arbitrary, and in particular large positive, then $x^2-4xy+y^2=0$. 
Remark: We can get more information by going to polar coordinates. Let $x=r\cos\theta$, $y=r\sin\theta$. Then our expression is equal to $r^2(1-2\sin 2\theta)$. 
Choosing $\sin 2\theta =\tfrac{1}{2}$ gives us $0$ for arbitrarily large $r$. But we can also find "directions" $\theta$, with $0\lt \theta\lt \tfrac{\pi}{2}$ that make $1-2\sin 2\theta$ positive, or negative. Then choosing large $r$ we can make our expression arbitrarily large positive, or arbitrarily large negative. With a little more care, we can use the idea to show that for any given $a$, $M$, there exist $x$ and $y$ larger than $M$ such that $x^2-4xy+y^2=a$. 

