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 $\epsilon$?


  • 6
    $\begingroup$ If you can take $\delta$ only depending on $\epsilon$ and not on $x$, you get a (stronger) notion called uniform continuity. I think a smooth function $\mathbb{R}\rightarrow \mathbb{R}$ is uniformly continuous iff its derivative is bounded (in absolute value), which isn't true here. $\endgroup$ Nov 30 '10 at 8:37
  • 3
    $\begingroup$ Are you saying that I can indeed let $\delta$ depend on $x$, if I only wanted to show this is continuous, but not necessarily uniform continuous? $\endgroup$
    – yunone
    Nov 30 '10 at 8:41
  • 3
    $\begingroup$ Yup. This is how you show that $f$ is continuous at a particular point $x$. $\endgroup$ Nov 30 '10 at 8:48
  • 2
    $\begingroup$ Actually, as you say, you are not supposed to let $\delta$ depend on $x$, if you take the Wikipedia definition. But $\delta$ can depend on $c$. $\endgroup$
    – AgCl
    Nov 30 '10 at 10:52
  • 5
    $\begingroup$ @Fubini: There has been a lot of confusion here; when most people say "depend on $x$", they really mean "depend on $c$". Be careful there. You are correct that $\delta$ is not allowed to depend on $x$. $\endgroup$ Nov 30 '10 at 15:24

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 $a\gt 0$).

  • 2
    $\begingroup$ Thanks Arturo, all the information is starting to come together much more clearly. I've also noticed while poking around that people talk about a function being continuous at a point. Is showing that a function is continuous the same as showing that it is continuous at every point in its domain? I guess my point of confusion is that you show that $f(x)$ is continuous at arbitrary $c$; does this imply $f(x)$ is continuous even though $\delta$ is different for each point? I'm just a little unsure since I heard earlier that $\delta$ should be independent of $c$. $\endgroup$
    – yunone
    Nov 30 '10 at 20:14
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    $\begingroup$ @Fubini: Yes, by definition a function $f(x)$ "is continuous" if and only if $f(x)$ is continuous at $c$ for every $c$ in the domain of $f(x)$. And, again, yes, by showing that $f(x)$ is continuous at $c$ for every $c$, you show $f(x)$ is continuous. Going back to the definition, if we wrote it down this time we would say: "for every $c$ and every $\epsilon\gt 0$, there exists a $\delta\gt 0$ such that for all $x$, if $|x-c|\lt \delta$ then $|f(x)-f(c)|\lt \epsilon$." Again, the $\delta$ is allowed to depend on anything that precedes it, but on nothing that comes after it. $\endgroup$ Nov 30 '10 at 20:25
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    $\begingroup$ @Fubini: Remember that in general "For every $b$ there is an $a$" is not the same thing as "There is an $a$ for every $b$". For example, 'for every person, there is a woman who that person's mother' is not the same as 'there is a woman who is the mother of every person'. The order of the quantifiers matters. $\endgroup$ Nov 30 '10 at 20:36
  • $\begingroup$ Thanks again for a great answer and helpful comments. $\endgroup$
    – yunone
    Nov 30 '10 at 20:43

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.

  • $\begingroup$ Thanks for your answer Alex. I did some reading beforehand, and I knew that $f(x)=1/x$ is continuous but not uniform continuous. I'm not sure if you saw my comment below, but how would I then state the definition for continuity, but not uniform continuity necessarily? Also, is putting $\delta=|xc|\epsilon$ a sufficient proof to show that $f(x)=1/x$ is continuous? $\endgroup$
    – yunone
    Nov 30 '10 at 9:02
  • $\begingroup$ @Fubini Note that for given $\epsilon$ and $x$ you need to find a $\delta$ such that whenever $c$ has distance less than $\delta$ from $x$, the distance of $f(c)$ from $f(x)$ is less than $\delta$. So the game is: I throw $\epsilon$ and $x$ at you, you say "ok, here is your $\delta$." I say "Really? Let's check. Here is a $c$ such that $|c-x|<\delta$. So you are saying that I am guaranteed to have $|f(c)-f(x)|<\epsilon$? Blimey! You seem to be right! Let's try another $c$ close to $x$, just to be sure..." In other words, you cannot let $\delta$ depend on $c$. $\endgroup$
    – Alex B.
    Nov 30 '10 at 9:07
  • $\begingroup$ @Fubini For the concrete function at hand, as you correctly say, you want to achieve $|x-c|<|xc|\epsilon$ by taking $c$ close enough to $x$. So the question in choosing $\delta$ is: how close does $c$ have to be to $x$ for this inequality to be true? $\endgroup$
    – Alex B.
    Nov 30 '10 at 9:11
  • $\begingroup$ Hmm, I've been staring at it, but I'm not seeing any possibilities. At first I thought they should be within an $\epsilon$ distance of each other, but that fails if $|xc|\leq 1$. How can you possibly determine something like that if $|xc|$ could take values less than or greater to $1$? Is it perhaps something like $|x-c|<|c|/2$? $\endgroup$
    – yunone
    Nov 30 '10 at 9:23
  • 1
    $\begingroup$ Oh, I see. I'm sorry, I didn't look at wikipedia. So then, I swapped $x$ and $c$ in my answer. The first paragraph of the question still describes uniform continuity though. @Fubini That's precisely the tricky part of the exercise. $\endgroup$
    – Alex B.
    Nov 30 '10 at 11:02

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).

  • $\begingroup$ So just let $\delta=|xc|\epsilon$? I wasn't even aware I finished the proof! What if it is not actually the case that $|x-c|<|xc|\epsilon$? I feel like I was assuming what I wanted to prove before proving it. $\endgroup$
    – yunone
    Nov 30 '10 at 8:52
  • $\begingroup$ Your post is jumbled up. "delta depends on epsilon but NOT delta" is probably not what you wanted to say. Also, the OPs proof is in fact NOT ok, because he was trying to prove uniform continuity, which is not true for the function at hand. $\endgroup$
    – Alex B.
    Nov 30 '10 at 8:53
  • 1
    $\begingroup$ @Alex, I'm just trying to prove that it's continuous for now since I'm just starting out. If that is the case, is putting $\delta=|xc|\epsilon$ a sufficient proof for continuity? $\endgroup$
    – yunone
    Nov 30 '10 at 8:57
  • $\begingroup$ @Fubini, that comment of mine referred to the post of user3971. That's why the comment appears under his answer. Your question is fine. $\endgroup$
    – Alex B.
    Nov 30 '10 at 9:08
  • $\begingroup$ @user3971: in this case, the point of dependence is $c$, not $x$. You are checking continuity at $c$, so $\delta$ can depend on $c$. $\endgroup$ Nov 30 '10 at 15:04

The inversion function $f(x)=1/x$ is continuous at any $c \gt 0$.

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),
$1/x \lt 1/k$
$1/c \lt 1/k$.
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 $\frac{|x-c|}{|xc|}= |x-c|\frac{1}{xc} < \delta \frac{1}{k^2} \lt k^2 \varepsilon \frac{1}{k^2} = \varepsilon$.


$\begin{aligned}|x - c|<\delta&\implies -\delta< x–c <\delta\\&\implies c -\delta< x < c+\delta\end{aligned}$

$(c – \delta< x < c +\delta)\land(0 <\delta< c)\land(x>0)\implies(0 < c –\delta)$

$c –\delta < x < c +\delta$ and $0 < c –\delta$ imply $\begin{aligned}\frac1x &<\frac1{c –\delta}\implies \frac1{c·x}\\&= \frac1{c·x}\\&<\frac1{c·(c – \delta)}\\\implies\frac{|x – c|}{|c·x|}&=\left|\frac1x – \frac1c\right| \\&<\frac{|x – c|}{c·(c –\delta)}\\&< \frac{\delta}{c·(c – \delta)}\end{aligned}$

Let $\varepsilon=\frac{\delta}{c·(c – \delta)}$. This implies $\varepsilon·c^2 – c·\varepsilon·\delta =\delta\implies\varepsilon·c^2 = [1 + c·\varepsilon]·\delta\implies\delta=\frac{\varepsilon·c^2}{1 + c·\varepsilon}$

$\delta=\frac{\varepsilon·c^2}{1 + c·\varepsilon}$ and $\delta< c$ imply $\begin{aligned}\frac{\varepsilon·c^2}{1 + c·\varepsilon}& < c\\\implies\frac{c·\varepsilon}{1 + c·\varepsilon}& < 1\\\implies c·\varepsilon &< 1 + c·\varepsilon\\\implies 0 &< 1\end{aligned}$

Therefore, $0 <\varepsilon\implies 0 < \frac{c^2·\varepsilon}{1 + c·\varepsilon}$ for every such $\varepsilon$, and $\begin{aligned}|x – c|&<\frac{c^2·\varepsilon}{1 + c·\varepsilon}\\\implies\left|\frac1x–\frac1c\right| &<\varepsilon\end{aligned}$

for every such $\varepsilon$. Therefore, $\lim\limits_{x\to c}\frac1x=1\forall c>0$.

I have no idea how to type and render mathematical symbols here in stack exchange, so if anyone can edit my reply and fix it, then that would certainly make my answer far more useful.


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