# Determine if the function is uniformly continuous

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!

• 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 $1$. – Theo Bendit Dec 18 '18 at 14:06
• – idea Dec 18 '18 at 14:08

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 $$\lvert x-y\rvert<\delta$$.
• Do we need to prove that in a neighborhood of 1 we get $|f(x)-f(y)|\ge 1$? – argiriskar Dec 18 '18 at 14:11
• 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 $\bigl\lvert f(x)-f(y)\bigr\rvert\geqslant1$. – José Carlos Santos Dec 18 '18 at 14:15