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Determine if the function: $$f(x)=\frac{x}{(1-x)^2}$$ $\forall x>1$ is uniformly continuous.

I know that it is not. But I'm having trouble proving it. I reach to a point that i get this: $$\frac{|x-y||1-xy|}{(1-x)^2(1-y)^2} \le \delta|1-xy|$$ and I cannot continue.

Any help would be appreciated!

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    $\begingroup$ Try finding a Cauchy sequence $x_n$ that fails to map to a Cauchy sequence. Try thinking of a sequence $x_n$ that converges to $1$, but is strictly greater than $1$. $\endgroup$ – Theo Bendit Dec 18 '18 at 14:06
  • $\begingroup$ wolframalpha.com/input/?i=x%2F(1-x)%5E2 may help $\endgroup$ – idea Dec 18 '18 at 14:08
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You know that $\lim_{x\to1^+}f(x)=\infty$. So, for any $\delta>0$, you can find $x,y\in(1,1+\delta)$ such that $\bigl\lvert f(x)-f(y)\bigr\rvert\geqslant1$. But $\lvert x-y\rvert<\delta$.

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  • $\begingroup$ Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|\ge 1$? $\endgroup$ – argiriskar Dec 18 '18 at 14:11
  • $\begingroup$ I didn't say that. Actually, I don't even know what it means. What I wrote was that in $(1,1+\delta)$ there are elements $x$ and $y$ such that $\bigl\lvert f(x)-f(y)\bigr\rvert\geqslant1$. $\endgroup$ – José Carlos Santos Dec 18 '18 at 14:15

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