# Prove uniformly continuous $f(x)=1/x$

Consider the function $f : (0, 1] → R$ defined by $f(x) = 1/x$.

Prove that, for any $0 < r < 1$, f is uniformly continuous on $[r, 1]$.

I was trying to use the theorem if $f$ be continuous on [a,b] then $f$ is uniformly continuous. But I don't know how to relate and explain closed interval use the given $0< r < 1$.

• Do you see that $[r,1]$ is also a interval "like" $[a,b]$ and $f$ is continuous on $[r,1]$? – Arpit Kansal Mar 3 '17 at 3:29
• yes I know, but I don't know how to relate 0 < r < 1 – bingyan zhu Mar 3 '17 at 3:33
• Well once you fix a $r$ then $[r,1]$ is a compact interval and $f$ is continuous on $[r,1]$ as i said hence $f$ is uniformly continuous. – Arpit Kansal Mar 3 '17 at 3:34
• It is just the range of values of $r$ that you need to prove that $f$ is uniformly continuous over $[r;1]$ – Graham Kemp Mar 3 '17 at 3:34
• I totally understand, but the interval keeps change. so how to prove f on a variable interval continuous? – bingyan zhu Mar 3 '17 at 3:38

Let $\epsilon > 0$, you need to find a $\delta(\epsilon)$ such $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$
Let $x,y\in[r,1]$ then $|f(x)-f(y)| = |\frac1x-\frac1y|=|\frac{y-x}{xy}|\le|y-x|\frac1{r^2}=|x-y|\frac1{r^2}<\epsilon$
Let $\delta=\epsilon*r^2$, and that concludes your proof.
• Since the function is only defined from domain $(0,1]$ then no, however it is uniformly continuous for all $a,b > 0$ – Rab Mar 3 '17 at 22:56
• Why are you allowed to say $\frac{1}{xy} \leq \frac{1}{r^{2}}$ exactly? Edit: Nevermind, I get it - since $r$ is the lower bound then $\frac{1}{r}$ is the largest possible value for both $\frac{1}{x}$ and $\frac{1}{y}$. – Alena Gusakov May 1 '19 at 16:35