# Let $\alpha>0$. Show that the function $f(x)=1/x$ given by $f:[\alpha,\infty)\rightarrow \mathbb{R}$ is uniformly continuous.

I encountered this question on my textbook but I think that the function $$f(x)$$ is not uniformly continuous for all $$\alpha$$. For example let us say $$\alpha=10^{-10}$$ and fix $$\epsilon=1$$. Then $$\exists \delta>0$$ s.t. $$\forall x,y\in\mathbb{R}$$ with $$|x-y|<0$$, $$|\frac{1}{x}-\frac{1}{y}|<1$$. Now, pick $$x\in[10^{-9},10^{-8}]$$ with $$x<\delta$$ and set $$y=x/10$$. Then $$|x-y|=|\frac{9x}{10}|<\delta, \text{but}$$ $$\lvert \frac{1}{x}-\frac{1}{y}\rvert=|-\frac{9}{x}|>1\quad \text{since x\in[10^{-9},10^{-8}]}$$ However, this contradicts with $$f$$ being uniformly continuous on its domain.

• You can consider this like the order below: For fix $\epsilon=1$ then "there exists $\delta>0$". This $\delta$ is decided by $|1/x-1/y|<\epsilon=1$. When you calculate that $|1/x-1/y|=|x-y|/xy \leq |x-y|/\alpha^2$, now you can decide that $\delta=\epsilon\alpha^2$. So your "$x<\delta$" is wrong. More generally, you don't know what $\delta$ exactly is. it is just a abstract number.
Choose $$\epsilon\gt0$$. Then for any $$x,y\in[\alpha,\infty)$$ we have that, for $$\delta=\alpha^2\epsilon\gt0$$, \begin{align} |x-y|\lt\delta &\implies|x-y|\lt xy\epsilon\\ &\implies\frac{|x-y|}{xy}\lt\epsilon\\ &\implies\left|\frac{x-y}{xy}\right|\lt\epsilon\\ &\implies\left|\frac1y-\frac1x\right|\lt\epsilon\\ &\implies\left|\frac1x-\frac1y\right|\lt\epsilon\\ \end{align} as required.