Is $f(x)=1/x$ continuous on $(0,\infty)$? I've never actually done a delta-epsilon proof, so I thought I'd try my hand at one. I decided to try it out for $f(x)=1/x$. If I understand correctly from the wikipedia article, I want to show for any $\varepsilon>0$, there exists a $\delta>0$ such that if $|x-c|<\delta$, then $|f(x)-f(c)|<\varepsilon$.
Anyway, I noticed that I want something like $$|f(x)-f(c)|= \left| \frac{1}{x} - \frac{1}{c} \right|=\frac{|x-c|}{|xc|}<\varepsilon .$$ So $|x-c|<|xc|\varepsilon$, which looks similar to the fact that I want $|x-c|<\delta$. However, I've also heard that one is never supposed to let $\delta$ depend on $x$. 
Is this the right direction?
How would I use this information to find a corresponding $\delta$ for each $\epsilon$?   
Thanks!
 A: There have been some confusing comments regarding dependence on $x$ or on $c$, so let me try to put it all together.
You are correct that $\delta$ should not depend on $x$. However, when one is proving that $f(x)$ is continuous at $c$, then $\delta$ is allowed to depend on both $\epsilon$  and $c$. 
Remember the definition: $f(x)$ is continuous at $c$ if and only for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that for all $x$, if $|x-c|\lt \delta$, then $|f(x)-f(c)|\lt \epsilon$. 
Notice how the existence of $\delta$ is mentioned before $x$ ever comes into the picture? That's an indication that $\delta$ cannot depend on $x$. On the other hand, both $f(x)$, $\epsilon$, and $c$ occur before $\delta$, which means that, absent any indication to the contrary, $\delta$ is allowed to depend on $f(x)$ (obviously), on $\epsilon$, and on $c$.
So here, you cannot pick $\delta=|xc|\epsilon$, because that would make $\delta$ depend on $x$.
The way to get around it is to get rid of the dependence on $x$. The key here is that since we are trying to make sure everything works if $x$ is "close enough" to $c$, then we will also have that $|x|$ will be very close to $|c|$. So we should be able to to control that division by $x$ in the expression $\frac{|x-c|}{|xc|}$. 
How? Well, if a particular $\delta_0$ works, then any smaller one will work as well. So we can always shrink $\delta$ a bit more if needed. Wo the first thing we can note is that we can always require that $\delta$ be smaller than both $1$ and than $\frac{c}{2}$; that is, we will require $\delta\lt\min\{1,\frac{c}{2}\}$. Why $1$? Because then I know that $c-1\lt x \lt c+1$; if $c-1\gt 0$, then this means that $\frac{1}{c+1}\lt \frac{1}{x} \lt \frac{1}{c-1}$, so we can "control" the value of $\frac{1}{x}$. Why less than $\frac{c}{2}$? Just in case $c-1\lt 0$. So let $\mu=\min\{1,\frac{c}{2}\}$. Then we can conclude that $\frac{1}{x}\lt \frac{1}{c-\mu}$. (We could get away with simply putting $\delta\lt\frac{c}{2}$; restricting it to less than $1$ is a common practice, though, which is why I put it here).
So, by requiring that $\delta\lt\min\{1,\frac{c}{2}\}$, we guarantee that $\frac{1}{|x|}\lt \frac{1}{c-\mu}$ (remember that we are working on $(0,\infty)$). What do we gain by this? Well, look:
$$|f(x)-f(c)| = \left|\frac{1}{x}-\frac{1}{c}\right| = \left|\frac{c-x}{xc}\right| = |x-c|\frac{1}{c}\cdot\frac{1}{x} \leq |x-c|\frac{1}{c(c-\mu)}.$$
So if we also ask that $\delta\lt c(c-\mu)\epsilon$, then we have:
$$|f(x)-f(c)|\leq |x-c|\frac{1}{c(c-\mu)} \lt \frac{\delta}{c(c-\mu)} \lt \frac{c(c-\mu)\epsilon}{c(c-\mu)} = \epsilon$$
which is what we want!
So, in summary, what do we need? We need to make sure that $\delta$ is:


*

*Less than $1$;

*Less than $\frac{c}{2}$; (both of these to ensure $\frac{1}{x}\lt\frac{1}{c-\mu}$);

*Less than $c(c-\mu)\epsilon$, where $\mu=\min\{1,\frac{c}{2}\}$. 


So, for instance, we can just let $\delta = \frac{1}{2}\min(1,\frac{c}{2},c(c-\mu)\epsilon)$, where $\mu=\min\{1,\frac{c}{2}\}$. 
In general, if you can let your $\delta$ depend only on $f(x)$ and on $\epsilon$ but not $c$, then we say $f(x)$ is uniformly continuous. This is a stronger condition than continuity, and often very useful. $\frac{1}{x}$ is not uniformly continuous on $(0,\infty)$, however (though it is on $[a,\infty)$ for any $a\gt 0$). 
A: The statement in your first paragraph would be the definition of uniform continuity (I interpret what you wrote as "for all $\epsilon$, there exists a $\delta$ such that for all $x$..."). For a related question, see Intuition for uniform continuity of a function on $\mathbb{R}$. Uniform continuity implies continuity, but is strictly stronger, and in fact, the function at hand is continuous on $(0,\infty)$, but not uniformly continuous.
The statement of continuity would be that for all $\epsilon$ and all $x$ in the range, there exists a $\delta$ such that... So $\delta$ depends on $\epsilon$ and on $x$. Intuitively: the greater the slope at $x$, the closer you will have to get to $x$ for the function values to be close. 
A: Your proof looks ok.  Delta can depend on epsilon and on c in the definition of "continuity".  A function is called uniformly continuous if you can prove that given epsilon, the required value of delta depends on epsilon but NOT c.  1/x is NOT uniformly continuous on (0,1) which is why you can't get around the fact that delta depends on c.  1/x would be uniformly continuous on other intervals however (intervals where f' is bounded).
A: The inversion function $f(x)=1/x$ is continuous at any $c \gt 0$. 
Proof:
Recall first that for any positive numbers $u, v$ and $w$,
$u \lt v \implies 1/u \gt 1/v \; \;\; \;\; \;\; \;$ (1)

$u \lt v \implies uw \lt vw  \;\;\; \;\; \;\; \;\; \;$ (2)

Let $\varepsilon \gt 0$ be given.
Find a number $k \gt 0$ such that $k \lt c$.  
Let $\delta = min(k^2 \varepsilon,\, c-k)$ and consider any $x$ with $|x-c|<\delta$.
Note that by (1),  $1/x \lt 1/k$ and $1/c \lt 1/k$.
So, by applying (2) two times,
$\frac{1}{xc} = (1/x)(1/c) \lt (1/k)(1/c) \lt (1/k)(1/k) = \frac{1}{k^2}$
So $\frac{|x-c|}{|xc|}= |x-c|\frac{1}{xc} <  \delta \frac{1}{k^2} \lt  k^2 \varepsilon    \frac{1}{k^2} = \varepsilon$.
A: Yes, you have the right idea. Indeed, the starting point is to realize that $|\frac{1}{x} - \frac{1}{c}| = |\frac{c - x}{cx}| = \frac{|x - c|}{|cx|} \lt \epsilon$, which you already managed to do. Also, you know that $|x - c| \lt \delta$, hence $\frac{|x - c|}{|cx|} \lt \frac{\delta}{|cx|}$. Thus, the piece that your proof is missing is realizing that $\frac{1}{|cx|} \lt f(\delta, c)$, hence $\frac{|x - c|}{|cx|} \lt \delta f(\delta,c)$. You then let $\epsilon = \delta f(\delta, c)$, and ensure that for every $\epsilon \gt 0$, $\delta \gt 0$ holds. Hence, your question can be simplified to asking, how can you find what $f(\delta, c)$ is? Others have provided some answers as to how, but what I want to do is provide an alternative approach, an approach I find to be far more intuitive and natural, and does not require utilizing the $\min$ function, so I think this is less contrived.
The key idea is to go back to the inequality $|x - c| \lt \delta$, and utilize the reverse triangle inequality. You will have $||x| - |c|| \leq |x - c|$, so $||x| - |c|| \lt \delta$, which is equivalent to $-\delta \lt |x| - |c| \lt \delta$, which its equivalent to $|c| - \delta \lt |x| \lt |c| + \delta$. We are now very close to being able to find bound $\frac{1}{|x|}$ as we want, but we cannot apply the reciprocal function without ensuring that $|c| - \delta \gt 0$. So you can simply assume this (*), so that $|c| - \delta \lt |x| \lt |c| + \delta$ implies $\frac{1}{|c| + \delta} \lt \frac{1}{|x|} \lt \frac{1}{|c| - \delta}$, which implies $\frac{1}{|cx|} \lt \frac{1}{|c|(|c| - \delta)}$. In other words, $f(\delta, c) = \frac{1}{|c|(|c| - \delta)}$.
To complete the proof, you now only need to ensure that if $\epsilon = \frac{\delta}{|c|(|c| - \delta)} \gt 0$, then $\delta \gt 0$. This is very simple. Because of the assumption (*), it follows that $f(\delta, c) = \frac{1}{|c|(|c| - \delta)} \gt 0$. This means that $$\epsilon = \delta f(\delta, c) \gt 0 \implies \delta \gt 0.$$
The proof is now complete.
