Why is $f(x)=\frac{1}{x}$ not uniformly continuous on $(0,1)$?

Question

I know that $f(x)=\frac{1}{x}$ is continuous on $(0,1)$ but why is $f(x)=\frac{1}{x}$ not uniformly continuous on $(0,1)$?

Answer 1

No matter which $\delta>0$ you choose, you can make $|\frac1x-\frac1{x+\delta}|$ arbitarily large by picking $x$ close enough to $0$.

Therefore there is no $\varepsilon$ that has a matching $\delta$ -- and uniform continuity requires that every $\varepsilon>0$ must have one.

Answer 2

Reason 1: $x_n= \frac {1}{n}$ is cauchy while $f(x_n)=n$ is not Cauchy.

Reason 2: lim$_{x \to 0^+} \frac {1}{x}$ does not exist.

Answer 3

Take $$ x_n=\frac{1}{n},\,y_n=\frac{1}{2n}. $$ Then $$ x_n-y_n\to 0\quad\text{while}\quad f(y_n)-f(x_n)=n\to\infty. $$

Note. Suppose $f : A\to\mathbb R$ is uniformly continuous. In such case, whenever $\{x_n\},\{y_n\}\subset A$ and $x_n-y_n\to 0$, then $f(x_n)-f(y_n)\to 0$ as well.