Proof of uniform continuity of $\frac{1}{x}$ 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. 
 A: Since $0\notin(0,\infty)$, $0$ is completely irrelevant to the question of whether $f$ is uniformly continuous on $(0,\infty)$. To show that $f$ is not uniformly continuous on $(0,\infty)$, you should show that there is some $\epsilon>0$ such that no matter what $\delta>0$ you pick, you can find points $x,y\in(0,\infty)$ such that $|x-y|\le\delta$, but $|f(x)-f(y)|>\epsilon$. HINT: You can take $\epsilon=1/2$. Now consider values of $x$ of the form $\frac1n$ for $n\in\Bbb Z^+$.
To prove that $f$ is uniformly continuous on $(\mu,\infty)$ for $\mu>0$, you need what is really the key insight for both questions: for $x>0$, the graph of $y=\frac1x$ gets steeper and steeper as $x$ gets smaller and smaller. Given $x,y\in(\mu,\infty)$ with $|x-y|\le\delta$, where $\delta$ is some as yet unspecified positive real number, can you find an upper bound on $|f(x)-f(y)|$? How does it compare with $\frac{\delta}{\mu^2}$? (Consider $f'(x)$.)
A: Notice that:


*

*$f$ is $C^1$.

*$f'$ is bounded away from $0$.

*If a $C^1$ function (on a connected subset of $\bf R$) has a bounded derivative, it is Lipschitz (this follows from fundamental theorem of calculus).

*Any Lipschitz function is uniformly continuous.


Alternatively, you can elementarily show that $f$ is Lipschitz on $(\mu,+\infty)$ for $\mu>0$, using the fact that $\lvert 1/x-1/y\rvert=\lvert x-y\rvert/(xy)^2$ and $xy>\mu^2$.
To show that $f$ is not continuous near zero, best way would be to prove it by contradiction, but the essential thing is that near zero, we get arbitrarily large jumps in arbitrarily small intervals.
A: To prove that $1/x$ is uniformly continuous on $(\mu,\infty)$, we can use the mean value theorem. Suppose $x,y>\mu$. Then, there is a $c$ strictly between $x$ and $y$ such that
$$
\frac{1}{x}-\frac{1}{y}=\frac{1}{c^2}\left(y-x\right) \, .
$$
Since $c>\min(x,y)$, we have $c>\mu$ and thus
$$
\left|\frac{1}{c^2}(y-x)\right|<\left|\frac{1}{\mu^2}(y-x)\right|=\frac{|y-x|}{\mu^2} \, .
$$
So given $\varepsilon>0$, set $\delta=\varepsilon\mu^2$.
