Show that the function $f(x) = \frac{1}{x}$ is not uniformly continuous on the interval $(0,\infty)$ but is uniformly continuous on any interval of the form $(\mu, \infty)$ if $\mu > 0$.
My Work
Referring to the definition of uniform continuity, I have that $f$ is unif. cts. if for each $\epsilon > 0$ there is a $\delta > 0$ so that for all $x, c$ in the domain of $f$ $|x - c| \le \delta \ \Rightarrow \ |f(x) - f(c) | \le \epsilon$.
From this definition, it is clear that if $f$ is uniformly continuous, it will be uniformly continuous on its domain, $(-\infty, \infty) \backslash \{0\}$. So for $\mu > 0$, $(\mu, \infty) \subset \mathrm{Dom}\,(f)$. Additionally, $f$ cannot be unif. cts on $(0, \infty)$ because $0 \notin \mathrm{Dom}\, (f)$. (Sorry about the longwindedness)
Now to find the $\delta$:
\begin{align*} |f(x) - f(c)| = \left|\frac{1}{x} - \frac{1}{c}\right| &= \left|\frac{x - c}{cx}\right| \\ \text{since }x\text{ is within }\delta\text{ of }c \ \Rightarrow \ &\le \frac{\delta}{|cx|}\\ x, c>0 \ \Rightarrow \ &= \frac{\delta}{cx} \end{align*}
This is where I am stuck. Should I use that $x \le c + \delta$, or should I break this up into two cases, one where $cx < 1$ and one where $cx \ge 1$?
Edit (due to Brian M. Scott)
It was pointed out that $0 \notin (0, \infty)$ so my above argument is senseless.