Is $f(x)=1/x$ continuous on $(0,\infty)$?

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 $a\gt 0$).

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).

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.