Continuity of product function

Let f(x,y)=xy
This function is continuous, as it is easy to verify using "Continuous Function on metric space"previous criteria.
Let $ (a,b)\in R^2 $ and let $ \epsilon >0 $ for all $ (x,y)\in R^2 $ such as $ \sqrt{(x-a)^2+(y-b)^2}<\delta $

we have
$ |xy-ab|=|xy-ay+ay-ab|\leq|y||x-a|+|a||y-b|\le\sqrt{y^2+a^2}\sqrt{(x-a)^2+(y-b)^2}<\sqrt{a^2+(1+|b|)^2}\delta' $

and the last product is less or equal to $ \epsilon $ if $ \sqrt{a^2+(1+|b|)^2}}\delta'\leq\epsilon $ So we can chose $ \delta=min\{1,\frac{\epsilon}{\sqrt{a^2+(1+|b|)^2)}}\} $ to have also $ |y|\leq|y-b|+|b|\leq\sqrt{(x-a)^2+(y-b)^2}+|b|<1+|b| $
used before.
As the choice of $ \delta $ depends on $ (a,b)\in R^2 $ the function is not uniform continuous on $ R^2 $ as it is easy to verify.

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