Proving a two variable limit using $\epsilon- \delta$ approach.

Question

I came across a question in which we had to prove the following limit by $\epsilon-\delta$ approach:

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

I had to prove that: For every $\epsilon>0$ in $\sqrt{(x+1)^2+(y+1)^2}<\epsilon$, there exists a corresponding $\delta$ such that $|xy-2x^2+1|< \delta$. So, I proceeded by putting $x+1=x_1$ and $y+1=y_1$ so that it would simplify the former expression inside the square root and would also help me to convert it to polar. However, neither of the methods worked and I ran out of ideas. I am new to multivariable calculus and am using $\epsilon-\delta$ approach in two variables for the first time.

Would someone please help me with this?

Answer 1

Take $x_1,y_1$ so small as $x_1^2+y_1^2<1$ \begin{align*} |xy-2x^2+1|&=|(x_1-1)(y_1-1)-2(x_1-1)^2+1|\\&=|x_1y_1-(x_1+y_1)+1-2x_1^2-2+4x_1+1|\\ &\le \color{blue}{|x_1||y_1|}+|x_1+y_1|+2x_1^2+4|x_1|\\ &\le \color {blue}{(\sqrt{x_1^2+y_1^2})^2}+6\sqrt {x_1^2+y_1^2}+2(x_1^2+y_1^2)\\&=\sqrt{x_1^2+y_1^2} (6+3\sqrt{x_1^2+y_1^2})\\&\lt 9\sqrt{x_1^2+y_1^2}=9\sqrt{(x+1)^2+(y+1)^2} \end{align*}

Can you take it from here?

Answer 2

\begin{align} xy-2x^2+1 &= (x+1-1)(y+1-1)-2(x+1-1)^2+1\\ &=(x+1)(y+1)-(x+1)-(y+1)-2(x+1)^2+4(x+1)\\ &=(x+1)(y+1)+3(x+1)-(y+1)-2(x+1)^2 \end{align}

We choose $\delta < 1$,

then we know that $$|xy-2x^2+1| \le \delta^2 + 3\delta + \delta + 2\delta = 7\delta$$

Now, I will leave the task of choosing $\delta$ as a function of $\epsilon$ to you.