Prove $\lim_{(x,y)\to(1,1)} x^2+xy+y=3$ 
Prove that $$\lim_{(x,y)\to(1,1)} x^2 + xy + y = 3$$ using the epsilon-delta definition.

What I have tried: 
Let $\epsilon > 0$ be arbitrary. We must show that for every $\epsilon$ we can find $\delta>0$ such that
$$0 < \|(x,y) - (1,1)\| < \delta \implies \|f(x,y) - 3\| < \epsilon$$.
Or equivalently, 
$$0 < \sqrt{(x-1)^2 + (y-1)^2} < \delta \implies |x^2+xy+y-3| < \epsilon$$
The problem is finding the $\delta$. I have been trying to manipulate $|x^2+xy+y-3|$ with no success. 
$|x^2+xy+y-3|$
$=|(x^2-1)+y(x+1)-2|$
$=|(x-1)(x+1)+y(x+1)-2|$
$=|(x+1)[(x-1)+y]-2|$
$=|(x+1)||[(x-1)+(y-1)-1|$
Any help is appreciated!
 A: \begin{align}
x^2+xy+y-3 &= (x-1)^2+2x-1+(x-1)(y-1)+x+y-1+(y-1)+1-3 \\
&=(x-1)^2+(x-1)(y-1)+(y-1)+3x+y-4 \\
&=(x-1)^2+(x-1)(y-1)+(y-1)+3(x-1)+(y-1)\\
&=(x-1)^2+(x-1)(y-1)+2(y-1)+3(x-1)\\
\end{align}
Let $\delta= \min(1, \frac{\epsilon}7),$
Then 
\begin{align}
|x^2+xy+y-3| &\leq |x-1|^2+|x-1||y-1|+2|y-1|+3|x-1| \\ 
&\leq 2\delta^2+5\delta \\
& \leq 7 \delta \\
& \leq \epsilon
\end{align}
Alternatively, let me work from where you left off, there is a mistake at the last line of your equation, it should be 
\begin{align}
(x+1)[(x-1)+(y)]-2 &=(x+1)[(x-1)+(y-1+1)]-2 \\
&=(x+1)[(x-1)+(y-1)]+x+1-2 \\
&=(x+1)[(x-1)+(y-1)]+(x-1)
\end{align}
Choose $\delta = \min(1, \frac{\epsilon}7)$
Then $|x-1|\leq \delta$ implies $1-\delta \leq x \leq 1+\delta$ and hence $|x+1|\leq 3,$
Hence,
 \begin{align}
|(x+1)[(x-1)+(y-1)]+(x-1)| &\leq 3 [|x-1|+|y-1|]+|x-1|\\
& \leq 3(2\delta)+\delta \\
& = 7\delta \\
& \leq \epsilon
\end{align}
A: It's easier if you substitute $x=t+1$ and $y=u+1$, so
$$
x^2+xy+y-3=(t+1)^2+(t+1)(u+1)+(u+1)-3=
t^2+tu+3t+2u
$$
Now, if $t^2+u^2<\delta^2$, which is the same as $\sqrt{(x-1)^2+(y-1)^2}<\delta$, we surely have $|t|<\delta$ and $|u|<\delta$, so
$$
|t^2+tu+3t+2u|\le
|t^2+u^2|+|t|\,|u|+3|t|+2|u|\le
2\delta^2+5\delta
$$
and we just need to keep $2\delta^2+5\delta<\varepsilon$, which is attained for
$$
0<\delta<\frac{-5+\sqrt{25+8\varepsilon}}{4}
$$
