<s>[INST] To show the $\epsilon-\delta$ definition for limits holds. Question: Check if the following limit exists, if so show that the $\epsilon$ $\delta$ definition for limits holds.
$$\lim_{(x,y) \to (1,2)} \frac{(x-1)^2(y-2)^2}{x^2+y^2-2xy-4y+5}$$
My answer: So clearly the limit is zero, you can write $\frac{(x-1)^2(y-2)^2}{x^2+y^2-2xy-4y+5}$ as $\frac{(x-1)^2(y-2)^2}{(x-1)^2+(y-2)^2+(1+2x-2xy)}$ and $\lim_{(x,y) \to (1,2)} \frac{1}{(y-2)^2}+\frac{1}{(x-1)^2}+\frac{(1+2x-2xy)}{(x-1)^2(y-2)^2}=0$. 
I am having difficulty showing the $\epsilon$-$\delta$ definition holds. The trouble I am having is bounding the denominator in $\frac{(x-1)^2(y-2)^2}{x^2+y^2-2xy-4y+5}$, the numerator can be handled quite easily.  
 [/INST] Hint:
Note we can factorize the denominator $D$ as
$x^2+y^2-2xy-4y+5=(x-\frac{y}{2})^2+\frac{3}{4}(y-2)^2-(y-2)-xy$
If $(x,y)$ closed to $(1,2)$, we know $(x-\frac{y}{2})^2,\frac{3}{4}(y-2)^2,(y-2)$ are very closed to $0$ and $-xy$ is closed to $-2$, hence we know the denominator $D$ satisfies
$$|D|\ge |xy|-|(x-\frac{y}{2})^2+\frac{3}{4}(y-2)^2-(y-2)|\ge 1,$$
when $(x,y)$ closed to $(1,2)$.
Then combine with numerator, which is closed to $0$ when $(x,y)$ closed to $(1,2)$, you can prove the limit is $0$ by translating the idea into $\epsilon-\delta$ language .
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