Prove $\lim_{(x,y)\to(1,1)} x^2+xy+y=3$

Question

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!

Answer 1

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 $$ 0<\delta<\frac{-5+\sqrt{25+8\varepsilon}}{4} $$

Answer 2

\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}