Proof of multi-variable limit

Question

So I am trying to show that $$\lim_{(x,y)\rightarrow(1,1)}\frac{2xy}{x^2+y^2}=1 $$ I seem to be doing my scratch work(in order to find $\delta$) wrong. Is this correct? \begin{align} \left|\frac{2xy}{x^2+y^2}-1\right | &= \frac{2\left|x\right |\left|y\right |}{x^2+y^2}-1 &\leq \frac{2\sqrt{x^2+y^2}\sqrt{x^2+y^2}}{x^2+y^2} -1 &= 2-1=1 < \epsilon \end{align} If not, then can someone please provide with a full-proof of how this will be tackled and what are we to take $\delta$ as including the corresponding scratch work?

Answer 1

I wrote what follows, first, and then came to the conclusion that it is totally overkill.

The numerator is continuous at $(1,1).$ The denominator is non-zero and continuous at the same point. Therefore $f(x,y)$ is continuous at $(1,1).$

$\lim_\limits{(x,y)\to(1,1)} f(x,y) = f(x,y)$

And if you want to kill it.

$\forall \epsilon>0, \exists \delta>0: d\big( (x,y),(1,1)\big)<\delta \implies \left|\frac{2xy}{x^2+y^2}-1\right |<\epsilon$

$\left|\frac{2xy}{x^2+y^2}-1\right |\\ \left|\frac {x^2 - 2xy + y^2}{x^2+y^2}\right |$

We need to show that:

$x^2 - 2xy + y^2 \le k\delta$

For some bounded k. We can choose the distance metric which is most convenient to use.

And that $x^2+y^2>0$ with an appropriately bounded $\delta$

Suppose $d\big( (x,y),(a,b)\big) = |x-a|+ |y-b|$ I will leave it to you to show that this is a valid metric.

$x^2 - 2xy + y^2 = (x-y)^2 = \big((x-1) - (y-1)\big)^2$

$d\big( (x,y),(1,1)\big) < \delta \implies ((x-1) + (y-1))^2<\delta^2$

Let $\delta \le 1$

$\frac 14 \le x^2+y^2 \le 5$

let $\delta = \min (1, \frac {\epsilon}{4})$