Determine whether $\lim\limits_{x \rightarrow x_0} f(x) = \pm\infty$ with $ f(x)=\frac{4y^2-x^2}{(x-2y)^3},x_0=(2,1)$ 
Determine whether $\lim\limits_{x \rightarrow x_0} f(x) = \pm\infty$ with $$ f(x)=\frac{4y^2-x^2}{(x-2y)^3}, x_0=(2,1)$$

By definition, $\lim\limits_{x \rightarrow x_0} f(x) = \infty$ if:


*

*$x_0$ is a limit point of $D_f$ 

*For every $M \in \mathbb R, \exists \delta>0$ s.t. $f(x)>M$ , $0<|x-x_0|<\delta$


.
But, $\lim\limits_{(x,y) \rightarrow (2,1)} f(x,y)$ does not exists as $(x-2y)^3 \rightarrow 0$.
I was trying to manipulate $f(x)$ to obtain an inequality placing in relation
$f(x)=\frac{4y^2-x^2}{(x-2y)^3}$   and $|x-x_0|$ 
Also, after Reducing 
$f(x)=\frac{4y^2-x^2}{(x-2y)^3} = - \frac{x+2y}{(x-2y)^2}$
I was trying to apply Schwarz's Inequality (i.e. $|x \cdot y| \leq |x||y|$).
I need some help to understand how to bound $f(x)$ and reach to inequalities with $M$ and $x-x_0$.
Much appreciated
 A: Note that 
$$f(x,y)=\frac{4y^2-x^2}{(x-2y)^3}=\frac{-(x+2y)(x-2y)}{(x-2y)^3}=\frac{-(x+2y)}{(x-2y)^2}.$$ When $(x,y)\to (2,1),$  The numerator goes to $-4$ and the denominator approaches $0$ from the right side. Hence the limit is $-\infty.$
$\textbf{EDIT:}$ Here is a solution with $\epsilon-\delta$ arguments, as requested on the comments.
Let $\epsilon >0.$ Choose $\delta < \frac{4}{3}$ and such that $3\epsilon \delta^2 +\delta <\frac{4}{3}$(Obviously this is always possible). Now, 
$$|x-2|<\delta,\;|y-1|<\delta$$ will imply
$$(x-2y)^2= |x-2y|^2=|x-2+2-2y|^2\leq (|x-2|+2|1-y|)^2\leq (\delta +2\delta)^2= 9\delta^2.$$ On the other hand, we have
$$x+2y\geq 2-\delta + 2(1-\delta)= 4-3\delta>0,$$  where the last inequality follows from definition of $\delta.$ hence, we will have 
$$f(x,y)=\frac{-(x+2y)}{(x-2y)^2}\leq \frac{3\delta-4}{9\delta^2}<-\epsilon,$$ as desired.
A: Note that $f(x,y)=\frac{-(2y+x)}{(x-2y)^2}$
The limit is $-\infty$
From the definition $\forall M>0$ exists $\delta>0$ such that $f(x,y)<-M$ for all $(x,y): \sqrt{(x-2)^2+(y-1)^2}< \delta$
Take $\delta <1$
So solving the inequality we have that $$|x-2|< \delta$$ $$|y-1|< \delta \Rightarrow x<2+ \delta,y<1+\delta$$
We can prove using the aboce two inequalities:
$$\frac{1}{(x-2y)} \geq \frac{1}{3(\delta+2)^2+6(\delta+1)^2}> \frac{1}{39}$$
when $\delta <1$
Also $$-(2y+x)<-[(2(\delta+1)+\delta+2]=-(4+3 \delta)<-3 \delta$$
Combining all the above we have that $$f(x,y)<\frac{-3\delta}{39}$$
Take $\delta<\min\{1,\frac{39M}{3}\}$
A: For every real $M>0$, $\exists \delta>0$ s.t. if $0<|(x,y)-(2,1)|<\delta$ then we let the neighborhood too small such that $1<x<3$ and $0<y<2$. Thus
$$|x-2|<\delta~~~,~~~|y-1|<\delta$$
from $f(x)=-\dfrac{x+2y}{(x-2y)^2}$ we have
\begin{align}
-f(x) =& \dfrac{x+2y}{(x-2y)^2}\\
>& \dfrac{1}{(x-2-2y+2)^2}\\
>& \dfrac{1}{9\delta^2}\\
=& M
\end{align}
