Is $f(x) = \frac{1}{x}$ uniformly continuous on $(1, \infty)$?

Question

is the function $f(x) = 1/x$ uniformly continuous on $(1, \infty)?$ I know that it is not uniformly continuous on $(0, \infty)$, but now I'm restricting it even more to get rid of most of the bad parts. I think the answer is yes.

Answer 1

If $x,y\in(1,\infty)$, then$$\left\lvert\frac1x-\frac1y\right\rvert=\frac{\lvert x-y\rvert}{xy}<\lvert x-y\rvert.$$So, for each $\varepsilon>0$, just take $\delta=\varepsilon$ and$$\lvert x-y\rvert<\delta\implies\left\lvert\frac1x-\frac1y\right\rvert<\varepsilon.$$

Answer 2

$f$ is differentiable at $(1,+\infty)$ and

$$|f'(x)|=|\frac{-1}{x^2}|\le 1$$

$$f' \text{ is bounded } \implies$$ $$f \text{ is Lipschitz} \implies$$ $$f \text{ is uniformly continuous}.$$