How can I prove this function is uniformly continuous?

Question

The function is $$ f : B((0,0),1) \rightarrow \Bbb R,\,f(x,y)={xy\over{x-1}}. $$ I tried with sequences: Let be $ \{(x_k,y_k)\} $ and $ \{(x'_k,y'_k)\} $ sequences of $B((0,0),1)$ such that $ \{(x_k-x'_k,y_k-y'_k)\}\rightarrow(0,0). $

So, $ \{f(x_k,y_k)- f(x'_k,y'_k)\}={x_kx'_k(y_k-y'_k)+x'_ky'_k-x_ky_k\over{x_kx'_k-x_k-x'_k+1}} $

I know $x_kx'_k(y_k-y'_k) $ tends to zero but I don't know what to do with $x'_ky'_k-x_ky_k.$

I previously tried with some sequences to find one such that $ \{f(x_k,y_k)- f(x'_k,y'_k)\}$ tends to a non-zero number so that the function is not uniformly continuous but I didn't find it.

Thanks!

Answer 1

Consider two sequences $(x_n^1,y_n^1)=(1-\frac{1}{n},\frac{1}{2})$, $(x_n^2,y_n^2)=(1-\frac{1}{2n},\frac{1}{2})$.

Note $\|(x_n^1,y_n^1)-(x_n^2,y_n^2)\|=\frac{1}{2n}\to 0$. But $\|f(x_n^1,y_n^1)-f(x_n^2,y_n^2)\|=\frac{n}{2}\to \infty$.

So it's not uniformly continuous.