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!

  • 1
    $\begingroup$ 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

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 $\lvert x-y\rvert<\delta$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.