Prove that $\frac{1}{x-1}$ is not uniformly continuous on (0,1)

Question

I'm solving this problem $\frac{1}{x-1}$ to show that it is not uniformly continuous on $(0,1)$. Here is my former trial on determining if it was;

$$\Big|f(x)-f(y)\Big|=\Big|\frac{1}{x-1}-\frac{1}{y-1}\Big|=\Big|\frac{y-x}{(x-1)(y-1)}\Big|=\frac{\big|x-y\big|}{(x-1)(y-1)}.$$

But $(0,1)$ didn't work, so I agree with Przemysław Scherwentke and José Carlos Santos, that it is not uniformly continuous on $(0,1)$. Can anyone help me show that it is not uniformly continuous?

Answer 1

Consider the sequence $x_n = 1-\frac 1{2^n}$

For all $n>1, x_n\in(0,1)$

For any $\epsilon$ and $\delta$, there exist $n$ such that that $|x_{n+1} - x_n| = \frac {1}{2^{n+2}} < \delta$ and $|f(x_n+1) - f(x_n)| = 2^{n} > \epsilon$

Answer 2

HINT: $f(x)=\frac{1}{x-1}$ has got a vertical asymptote at 1. Can $f$ be uniformly continuous?

Extension. Let $x_n=1-1/2^n$. It is a Cauchy sequence. But $f(x_n)$ is not Cauchy, because it tends to infinity (even the difference of two consecutive terms).

Answer 3

The sequences $x_n=1-\frac{1}{n\pi}$ and $y_n=1-\frac{1}{(n+1)\pi},\;n\in \Bbb{N}$ will work, since $x_n,y_n\in (0,1)$. So, letting $\delta>\frac{1}{n},\;n\in \Bbb{N}$

$$\big|x_n-y_n\big|<\delta.$$ But $$\big|f(x_n)-f(y_n)\big|>\pi.$$