$P:\mathbb{R}^2 \to \mathbb{R}, P(x,y) = x.y$ is continuous. I need to prove $P:\mathbb{R}^2 \to \mathbb{R}, P(x,y) = x.y$ is continuous.
I just proved that the sum is continuous, but I'm lost at how to manipulate the inequalities in this case.
My attempt:
Let $\epsilon>0$.
$d((x,y),(a,b)) = \sqrt{(x-a)^2 + (y-b)^2} $. So I need to find some $\delta$ such that $\sqrt {(x-a)^2 + (y-b)^2} < \delta \Rightarrow |xy-ab| < \epsilon $.
I verified that $\sqrt{(x-a)^2 + (y-b)^2} \geq |x-a| $ and $\sqrt{(x-a)^2 + (y-b)^2} \geq |y-b|$, but can't see if that helps. I don't know how can I find $\delta$ in this case, can someone give me a hint?
Thanks.

Since
$$
\underbrace{(x_1^2+x_2^2)}_{\|x\|^2}\underbrace{(y_1^2+y_2^2)}_{\|y\|^2}=\underbrace{(x_1y_1+x_2y_2)^2}_{(x\cdot y)^2}+(x_1y_2-x_2y_1)^2
$$
We have
$$
|x\cdot y|\le\|x\|\,\|y\|
$$
Thus,
$$
\begin{align}
\left|\,x\cdot y-a\cdot b\,\right|
&=\left|\,(x-a)\cdot b+x\cdot(y-b)\,\right|\\
&\le\|x-a\|\,\|b\|+\|y-b\|\,\|x\|
\end{align}
$$
If we have $\|x-a\|\le\min\left(\frac{\epsilon}{2\|b\|},\frac{\|a\|}2\right)$ and $\|y-b\|\le\frac{\epsilon}{3\|a\|}\le\frac{\epsilon}{2\|x\|}$, then $\left|\,x\cdot y-a\cdot b\,\right|\le\epsilon$.
Therefore, if we let $\delta=\min\left(\frac{\epsilon}{2\|b\|},\frac{\|a\|}2,\frac{\epsilon}{3\|a\|}\right)$, and $\sqrt{\|x-a\|^2+\|y-b\|^2}\le\delta$, then $\left|\,x\cdot y-a\cdot b\,\right|\le\epsilon$.
