Discontinuity: Point discontinuity

[samir]

A function that is continuous is a function whose graph has no breaks in it; i.e. it is a continuous curve. Generally speaking, a function is continuous if you can draw its graph without picking up your pencil.

At the very basic level, I understand this notion of discontinuity. But I am looking to understand this better, at a deeper level. Because I know I will need to know this well as I start to explore piece-wise functions.

"Did you lift your pencil while drawing the graph", the teacher asked.
"Yes?... I'm sure I did at some point", the student responded.
"There you go then! It's discontinuous!" The teacher turned around and went away.

Why do we bother drawing graphs? I want to be able to tell if a function is discontinuous without using a graph. Can't we tell if it is discontinuous unless we draw a graph? It is very unsettling to know that my pencil is smarter than me! (Nerd) Perhaps the next time I'm in doubt I will ask my pencil for the correct answer.

---

The simplest kind of discontinuity might be this thing called "point discontinuity". So let's start with that.

The following function is given.

$$\displaystyle f(x)=\begin{cases}1, & x=3 \\x^{2}, & \text{all other real x values} \\ \end{cases}$$

Is this function defined or undefined? Is it well-defined or ill-defined? Is it continuous or discontinuous?

If a function is discontinuous at a single point, does that mean that the function is also undefined at that point? Would that depend on the domain and rule of the function? What I'm having hard time with is understanding the difference between a function that's undefined at a point and a function that's discontinuous at a point. Can a function be discontinuous at a point and still be defined at that point? Can it be both discontinuous and undefined at a point? Perhaps a Venn diagram or some other diagram would tell more than words do alone. But let's start with a graph of the function above.

View attachment 5460

The point $f(3)=1$ is part of the function definition. So do we say that this function is defined at $x=3$? Do we say that it is defined but discontinuous at $x=3$?

Is there a difference in having a function that's undefined at a point, and having a function that's not a function at all (but a relation)? A function that is undefined at a point is interpreted as if the function doesn't exist at that point?

How do we tell which part of the function rule to use in the example above for $x=3$?

$$3 \to 1$$

$$x^2 | x=3 \to 3^2 \to 9$$

$$f(3)=9 \text{ or } f(3)=1$$

Which one is it? How do we decide between the two? My calculator says 1! But why?

I know this is not true:

$$f(3)=9 \text{ and } f(3)=1$$

We can't have two different output values for the same input value. So what does this tell us? The function doesn't exist at $x=3$?

Is the function discontinuous in its domain? It appears to be defined for all real numbers.

What if we limit the domain of the function to an interval before or after the $x=3$ point?

$$Domain = \left\{x \in\Bbb{R} | x \gt 3\right\}$$

$$Codomain = \left\{y \in\Bbb{R} | y \gt 9\right\}$$

Alternatively:

$$Domain = \left\{x \in\Bbb{R} | x \lt 3\right\}$$

$$Codomain = \left\{y \in\Bbb{R} | y \lt 9\right\}$$

Would it be continuous in one of these domains? Despite what it says in the function rule?

---

Let's have a look at this one. This function is only slightly changed. The point discontinuity in this case overlaps the graph of the function. As can be seen by the graph below.

$$\displaystyle f(x)=\begin{cases}9, & x=3 \\x^{2}, & \text{all other real x values} \\ \end{cases}$$

View attachment 5461

At $x=3$ the function value is 9. But at the same time, the function value for the point is also 9.

$$3 \to 9$$

$$x^2 | x=3 \to 3^2 \to 9$$

$$f(3)=9 \text{ or } f(3)=9$$

$$f(3)=9 \text{ and } f(3)=9$$

In this example, either of the two pieces of the function rule can be used to arrive at the same function value. So this function is continuous then?

Both in this example, and the previous example, I would refer to these points as points of inclusion. That is to say they are in the domain of real numbers. It doesn't say anything in the definition what the input value is not allowed to be. It's more as a consequence of the function rule that we might say that certain input values are not appropriate?

---

Here is another example, where we have a point of exclusion. That is to say, the point needs to be excluded from the domain by the very definition. We are not allowed to divide by zero.

$$f(x)=\frac{x^2(x-2)}{x-2}$$

$$f(2)=\frac{2^2(2-2)}{2-2}$$

$$f(2)=\frac{4(0)}{0}$$

$$f(2)=\frac{0}{0}$$

The function doesn't exist at $x=2$? But it does exist at all other points?

The function doesn't exist at all, at any x if its domain is defined to be all real numbers? It takes only one point in the domain at which the function is undefined for the entire function to be undefined or not to exist at all? What does that say about discontinuity? Do we even have to consider discontinuity if the function doesn't exist?

What if we changed the domain so that it excludes the point that breaks the universe (division by zero)?

If $x=2$ is not allowed, we can simplify the function rule.

$$f(x)=\frac{x^2(x-2)}{x-2}$$

$$f(x)=x^2$$

$$\displaystyle f(x)=\begin{cases}x^2, & x\neq 2 \\undefined, & x=2 \\ \end{cases}$$

View attachment 5463

What are the criteria for a function to be discontinuous at a single point? What methods do we have to determine if it is discontinuous at a point? Can we examine the limits at the given point do determine if the function is discontinuous at that point? How would the limit need to be?

 

[samir]

Here is a simple linear function with one exclusion point.

$$f(x)=2x, x\neq 2$$

We can rewrite this as:

$$\displaystyle f(x)=\begin{cases}2x, & x \in \Bbb{R} \\ undefined, & x=2\end{cases}$$

View attachment 5467

As with the previous examples, this function is undefined at $x=2$. So for this reason the function is discontinuous at that point? But the function is defined and exists at all other points?

What about intervals?

$$\displaystyle f(x)=\begin{cases}2x, & x \lt -1 \text{ and } x \gt 1 \\ undefined, & x \geq -1 \text{ and } x \leq 1 \\ \end{cases}$$

View attachment 5468

Would this function be discontinuous in the interval where it is undefined?

 

[Deveno]

That's a lot of ground to cover in one post, so I'll only hit some high-lights, and maybe we can fill in the gaps later.

The easiest "naive" way I can think of to describe continuity is this:

If $x$ is near the number $a$, then $f(x)$ is near the number $f(a)$. This isn't much different than the "pencil" definition (no sudden "jumps" or "gaps").

Strictly speaking, though, this isn't true-it's only true for those functions $f$ which are defined "near $a$" and AT $a$.

The trouble with making this more precise is (as you may suspect) partly with the word "near", or as I like to put it: "how close is close enough to be called near?"

The situation is made even more complicated by something you may or may not have suspected: the nature of the real numbers. Let me give a brief example:

Here is a sequence of rational numbers (since I'm lazy I'll write these in decimal form):

$1$
$1.4$
$1.41$
$1.414$
$1.4142$
$1.41421$
$1.414214$
$1.4142136$
$1.41421356$
$1.414213562$

(I'll stop now).

It appears this sequence of rational numbers is getting "close to something", but what? (If you guessed $\sqrt{2}$ you win a prize). Well, it turns out that $\sqrt{2}$ is NOT a rational number (I can prove this if you like, but I'd rather keep going for now). So, evidently, the rational numbers aren't "big enough" to include things which "still act like numbers"-we can approximate these things, we just can't get "all the way there".

This suggests that if we were to lay the rational numbers out on a line, they would "appear" to be solid, but some stuff would "still slip through" (like a very fine cheesecloth still let's water through). But we intuitively expect a line to be "continual" (numbers should form a continuum).

So what do we do?

The answer (and it eluded some very smart people for thousands of years) is to identify numbers with "sequences that approximate them". There is some additional machinery that goes along with this, but essentially it allows "infinite decimals" to be regarded as "numbers", too, even if we can't write them "all the way down". We call this "expanded set of numbers" the LIMIT POINTS of the rationals. It turns out we need these to talk about limits of real numbers, for otherwise our limits (even if we're only considering rational numbers) will be "something unknown". In other words, we need a number system "rich enough" to be closed under the taking of limits. In analysis, this property is called "completeness", and it's essentially a SPATIAL, or GEOMETRIC property.

And, now, we have something we can begin to fashion into a "proper definition" of continuity, the concept of a limit. What we mean, intuitively, by:

$\lim\limits_{x \to a} f(x) = L$

is: "$f(x)$ gets very close to $L$, when $x$ gets close to $a$".

But, again, there is a problem with the vagueness of "close". So we need a couple of extra concepts, here: a way to "compare closeness", and a way to say "arbitrarily close (but perhaps not equal)". The old-fashioned term is "infinitesimally close".

Here is the way we compare closeness, the concept of *distance*. We DEFINE, for two real numbers $x$ and $y$, the distance between them to be:

$d(x,y) = |y - x|$.

Note that this agrees with our intuitive notion of "how far apart" two integers are; for example, $-3$ and $4$ are "$7$ apart", and indeed:

$d(-3,4) = |4 -(-3)| = |4+3| = |7| = 7$.

So, now we can "quantify closeness", if $\delta$ is a positive real number, if $|x - a| < \delta$, we say $x$ is "within $\delta$ of $a$. Obviously, by "close", then, we mean $\delta$ should be "small" (near 0). Well, how small is "small"? It turns out that "it depends".

On what, you might ask?

Well, we would like $f(x)$ to be "very close" to $L$. If we pick some (small) number $\epsilon > 0$ as our "acceptable error" (so that it's "close enough for our standards"), we'd like to pick $\delta$ small enough to "make this happen".

In formal symbols:

"Pick $\delta > 0$ so that $|x - a| < \delta \implies |f(x) - L| < \epsilon$.

Now if we can find a $\delta$ EVERY SINGLE TIME, no matter which $\epsilon$ our boss wants us to comply with (even a millionth of a millionth, or tinier), we say:

"The limit of $f$ as $x$ approaches $a$ is $L$."

Well, almost. There's one slight wrinkle, which I hope an example will illuminate:

Suppose $f(x) = \dfrac{x}{x}$.

As you can see, $f$ is undefined at $0$, but "everywhere else" it's $1$. It's seems reasonable that we should define limit in such a way that:

$\lim\limits_{x \to 0} \dfrac{x}{x} = 1$.

So we'll assume our evil taskmaster has chosen some really small (and top secret) $\epsilon$. We need to find $\delta$ so as to out-wit him. We don't know what $\epsilon$ is, so we'll have to express $\delta$ in terms of it (hopefully this will then work every time). Let's try $\delta = \epsilon$.

Now if $|x - 0| < \epsilon$, then:

$\left|\dfrac{x}{x} - 1\right| = \left|\dfrac{x}{x} - \dfrac{x}{x}\right| = 0 < \epsilon$, right?

Well, no...there's a problem: if $x = 0$ we have $|x| < \epsilon$, but then $f(x)$ is undefined, and we can't even evaluate:

$|f(x) - 1|$.

So what do we do? We modify our definition of limit so $x$ can get "almost to $a$" (in our example $a = 0$) but not "touch it", so here is our amended definition of limit:

For ANY $\epsilon > 0$ there is SOME $\delta > 0$ such that WHENEVER:

$0 < |x - a| < \delta$

$|f(x) - L| < \epsilon$.

And finally, we say:

$f(x)$ is continuous at $a$ if:

$\lim\limits_{x \to a} f(x) = f(a)$.

Note that this means $f(a)$ has to exists, so any function undefined at $a$ CANNOT be continuous at $a$.

 

[I like Serena]

Hey samir! (Smile)

Your opening post is indeed a lot to take in.
Let me try to address just some of the issues.

The function $f$ is defined simply because we have a definition for it.
It's well-defined because it has a proper domain, which is $\mathbb R$, and for every element of that domain, an image has been defined that is indeed part of the codomain.
It's not continuous because there is a jump.

For the record, continuity in a point requires:

1. That the function is defined in that point.
2. That however close we want to be to that point, we can find a (small enough) interval such that every point in it is that close.
   This is the concept of a limit as Deveno explained, which should otherwise be interpreted as that the pencil doesn't leave the paper.
   I'm assuming (and hoping) that this will be explained in more depth as you progress in your courses.

 

[samir]

The easiest "naive" way I can think of to describe continuity is this:

If $x$ is near the number $a$, then $f(x)$ is near the number $f(a)$.

Let me see if I understand this first part correctly.

$$f:\Bbb{R} \to \Bbb{R}$$

$$f(x)=2x$$

If $x=a$ is a number in the domain of the function, then we have the following limits.

$$\lim_{{x}\to{a^{-}}}f(x) = f(a)$$

$$\lim_{{x}\to{a^{+}}}f(x) = f(a)$$

Left limit is equal to the right limit.

View attachment 5474

This describes continuity at a point, not discontinuity.

What is it that makes this description (of continuity rather than discontinuity) naive?

This isn't much different than the "pencil" definition (no sudden "jumps" or "gaps").

I think this differs a lot from the pencil definition. Why? Because the pencil definition describes a rule for identifying discontinuity, not continuity. I understand that continuity and discontinuity are each others opposites. But that doesn't mean we can use the pencil definition (a rule for discontinuity) as an argument for describing continuity. At least I don't see how. I think the pencil definition applies only to something called "jump discontinuity" (which you also hinted at).

Strictly speaking, though, this isn't true-it's only true for those functions $f$ which are defined "near $a$" and AT $a$.

What you're saying here is that using limits to define continuity at a single point is good for functions where the point of interest is included in the domain of that function?... something I referred to as inclusion points earlier. An example would be the one I gave above with the linear function.

What about functions where we have exclusion points? Why can't we use limits to check continuity or discontinuity at a point for those functions?

The trouble with making this more precise is (as you may suspect) partly with the word "near", or as I like to put it: "how close is close enough to be called near?"

The situation is made even more complicated by something you may or may not have suspected: the nature of the real numbers.

Making what more precis, exactly? The closeness of the $x$ as it approaches $a$ given by the example above?

By nature of the real numbers, you are referring to infinite decimals? But what if we made it more simple?

$$f:\Bbb{Z} \to \Bbb{R}$$

$$f(x)=2x$$

$$\lim_{{x}\to{a^{-}}}f(x) = f(a)$$

$$\lim_{{x}\to{a^{+}}}f(x) = f(a)$$

Now you can't argue that those are not the limits as $x$ approaches $a$ in the example? Or at least, it should be harder to argue now than it was when the domain of the function was all real numbers. Why? Because there are fewer numbers now between $x=n\lt a$ and $x=a$ or $x=n\gt a$ and $x=a$ where $n$ is the number that approaches $a$ from either left or the right on the number line.

Here is a sequence of rational numbers (since I'm lazy I'll write these in decimal form):

$1$
$1.4$
$1.41$
$1.414$
$1.4142$
$1.41421$
$1.414214$
$1.4142136$
$1.41421356$
$1.414213562$

Why rational? (Would you have written them in expanded form if you had not been lazy that day?)

It appears this sequence of rational numbers is getting "close to something", but what? (If you guessed $\sqrt{2}$ you win a prize).

I was unintentionally scanning my mind for a number while reading the rest of the text, but it didn't jump out at me immediately as the square root of two.

Well, it turns out that $\sqrt{2}$ is NOT a rational number (I can prove this if you like, but I'd rather keep going for now).

I know it is irrational. Or so I have been taught. But I have never seen any formal proof for this. But is this really necessary for understand the concept of discontinuity at a point? If it's not I would rather leave that for another discussion.

Is it not enough to just use some infinite decimal to illustrate any point you may have? I mean how is an irrational number different from an infinite decimal? For all I know one has periodicity and one does not. But both types have infinite or non-terminating decimals. The only reason I can think of for picking an irrational number as an example is that it is easier to pick, easier than having to find some large infinite decimal and make sure it has no periodicity to ensure it is rational.

So, evidently, the rational numbers aren't "big enough" to include things which "still act like numbers"-we can approximate these things, we just can't get "all the way there".

Not big enough? Well of course not, you made sure of that, didn't you? If you had kept to the real number set you would not have had this problem now. The set of rational numbers can't hold irrational numbers. Obviously!... or it would not be called a set of rational numbers.

"Things"? You mean numbers? "Still act like numbers"? Well if they are not numbers, then what are they? Are you referring to irrational numbers here?

"We just can't get all the way there." If you're referring to infinity then it's understandable. It's hard to get to some place infinite. That's the idea. It's infinite... endless, unreachable!

This suggests that if we were to lay the rational numbers out on a line, they would "appear" to be solid, but some stuff would "still slip through" (like a very fine cheesecloth still let's water through). But we intuitively expect a line to be "continual" (numbers should form a continuum). So what do we do?

(I like cheese!) What stuff? You mean like the way many rational numbers slip through the continuum of integers? A number line is only a model for the continuum of numbers. To indicate that a number is to be excluded from the continuum — in other words the set — we draw circles on the line.

The answer (and it eluded some very smart people for thousands of years) is to identify numbers with "sequences that approximate them". There is some additional machinery that goes along with this, but essentially it allows "infinite decimals" to be regarded as "numbers", too, even if we can't write them "all the way down".

You mean numbers like these?

$$\frac{1}{3} \approx 1.33333 ...$$

$$\sqrt{2} \approx 1.414213562 ...$$

So both of these are infinite decimal numbers? One is rational and one is irrational. You would now need to show me the sequences that can approximate them.

We call this "expanded set of numbers" the LIMIT POINTS of the rationals. It turns out we need these to talk about limits of real numbers, for otherwise our limits (even if we're only considering rational numbers) will be "something unknown". In other words, we need a number system "rich enough" to be closed under the taking of limits. In analysis, this property is called "completeness", and it's essentially a SPATIAL, or GEOMETRIC property.

Are you referring to this thing called "Extended real number line"?

 

[samir]

And, now, we have something we can begin to fashion into a "proper definition" of continuity, the concept of a limit. What we mean, intuitively, by:

$\lim\limits_{x \to a} f(x) = L$

is: "$f(x)$ gets very close to $L$, when $x$ gets close to $a$".

Did you intentionally use "very close" and "close"? As to say that $f(x)$ gets very close to $L$ as $x$ gets only "close" to $a$. I would expect them to be equidistant. But that might depend on the function rule.

But, again, there is a problem with the vagueness of "close". So we need a couple of extra concepts, here: a way to "compare closeness", and a way to say "arbitrarily close (but perhaps not equal)". The old-fashioned term is "infinitesimally close".

By "compare closeness" you mean the absolute distance between two numbers?

What do you mean by "arbitrarily close"? Is this the same as "infinitesimally close"? Do you mean the distance between a fixed number and a positive or negative infinity?

Here is the way we compare closeness, the concept of *distance*. We DEFINE, for two real numbers $x$ and $y$, the distance between them to be:

$d(x,y) = |y - x|$.

Note that this agrees with our intuitive notion of "how far apart" two integers are; for example, $-3$ and $4$ are "$7$ apart", and indeed:

$d(-3,4) = |4 -(-3)| = |4+3| = |7| = 7$.

I understand this. Distance is the way to compare closeness between two numbers. But these examples show the difference or distance between two fixed numbers. What you want to introduce here is infinity as one of the endpoints?

So, now we can "quantify closeness", if $\delta$ is a positive real number, if $|x - a| < \delta$, we say $x$ is "within $\delta$ of $a$.

What is $x$? What is $a$? Are they real numbers? Is $a \lt x$? Is $a \gt x$? Is $x \lt \delta$? Is $a \lt \delta$? How can $x$ be the distance $\delta$ of $a$? Can you give an example?

$$\delta=13$$

$$x=7$$

$$a=5$$

$$|7-5|<13$$

$$|2|<13$$

$$2<13$$

So 7 is not the distance 13 of 5?

Obviously, by "close", then, we mean $\delta$ should be "small" (near 0). Well, how small is "small"? It turns out that "it depends".

On what, you might ask?

On our "standards" or "acceptable error"?

Well, we would like $f(x)$ to be "very close" to $L$. If we pick some (small) number $\epsilon > 0$ as our "acceptable error" (so that it's "close enough for our standards"), we'd like to pick $\delta$ small enough to "make this happen".

To make what happen? Make $f(x)$ very close to $L$ you mean?

In formal symbols:

"Pick $\delta > 0$ so that $|x - a| < \delta \implies |f(x) - L| < \epsilon$.

Now if we can find a $\delta$ EVERY SINGLE TIME, no matter which $\epsilon$ our boss wants us to comply with (even a millionth of a millionth, or tinier), we say:

"The limit of $f$ as $x$ approaches $a$ is $L$."

Well, almost. There's one slight wrinkle, which I hope an example will illuminate:

Suppose $f(x) = \dfrac{x}{x}$.

As you can see, $f$ is undefined at $0$, but "everywhere else" it's $1$. It's seems reasonable that we should define limit in such a way that:

$\lim\limits_{x \to 0} \dfrac{x}{x} = 1$.

So we'll assume our evil taskmaster has chosen some really small (and top secret) $\epsilon$. We need to find $\delta$ so as to out-wit him. We don't know what $\epsilon$ is, so we'll have to express $\delta$ in terms of it (hopefully this will then work every time). Let's try $\delta = \epsilon$.

Now if $|x - 0| < \epsilon$, then:

$\left|\dfrac{x}{x} - 1\right| = \left|\dfrac{x}{x} - \dfrac{x}{x}\right| = 0 < \epsilon$, right?

Well, no...there's a problem: if $x = 0$ we have $|x| < \epsilon$, but then $f(x)$ is undefined, and we can't even evaluate:

$|f(x) - 1|$.

So what do we do? We modify our definition of limit so $x$ can get "almost to $a$" (in our example $a = 0$) but not "touch it", so here is our amended definition of limit:

For ANY $\epsilon > 0$ there is SOME $\delta > 0$ such that WHENEVER:

$0 < |x - a| < \delta$

$|f(x) - L| < \epsilon$.

You completely lost me here! (Whew)

And finally, we say:

$f(x)$ is continuous at $a$ if:

$\lim\limits_{x \to a} f(x) = f(a)$.

Note that this means $f(a)$ has to exists, so any function undefined at $a$ CANNOT be continuous at $a$.

Finally! (Party)

 

[Deveno]

Well, you sort of nailed it: a function is discontinuous at a point, if it is NOT continuous at said point.
First, I'd like to give a proof a continuity at ALL points for your "very simple function" $f: \Bbb R \to \Bbb R$ given by $f(x) = 2x$, so you can see how such a proof usually goes.

Our goal, is to show for ALL real numbers $a$, that:

$\lim\limits_{x \to a} 2x = 2a$.

To do this, we must establish that, given any (usually assumed to be "arbitrarily small") $\epsilon > 0$ (here, $\epsilon$ has been chosen for us, so it is "some positive real number" which we can use in our proof, but we cannot make any assumptions about its size-it might be very close to 0), we can find SOME $\delta > 0$ such that:

$0 < |x - a| < \delta \implies |2x - 2a| < \epsilon$.

The main question in these sorts of investigations is "how to find $\delta$", and usually one discovers this by "working backwards" from where one "wants to get to".

Notice that $|2x - 2a| = |2(x - a)| = |2||x - a| = 2|x - a|$.

So if $\delta \leq \dfrac{\epsilon}{2}$, we have:

$0 < |x-a| < \delta \implies |x - a| < \dfrac{\epsilon}{2}$, and thus:

$|2x - 2a| = 2|x - a| < 2\dfrac{\epsilon}{2} = \epsilon$.

So, for the function $f(x) = 2x$, for any point $a$, we can pick $\delta \leq \dfrac{\epsilon}{2}$, which exists because $\epsilon$ does (for example if $\epsilon = \frac{1}{2}$, we could use any $\delta$ between 0 on up to (and including) $\frac{1}{4}$).

Note how this constrains the graph (or our "pencil" drawing it), if our point $(a,f(a)) = (a,2a)$ is on the graph, on the interval $(a - \frac{\epsilon}{2}, a + \frac{\epsilon}{2})$ the range is "trapped" within $(2a - \epsilon, 2a + \epsilon)$, no matter how close we zoom in, so when we "zoom all the way in", the function at $a$ has nowhere to go EXCEPT $2a$.

So, back to your questions about discontinuities:

If a function $f(x)$ (defined from $\Bbb R - \{a\} \to \Bbb R$) is defined "everywhere BUT $a$", but the limit:

$\lim\limits_{x \to a} f(x)$ EXISTS (and is equal to $L$), then the function $g: \Bbb R \to \Bbb R$:

$g(x) = f(x), x \neq a$
$g(a) = L$

then $g$ IS continuous at $a$ (since $\lim\limits_{x \to a} g(x) = \lim\limits_{x \to a} f(x) = L = g(a)$), and $f$ is said to have a "removable discontinuity" at $a$. You can think of this as "plugging a hole".

That leaves "jump discontinuties". I will illustrate with an example of how our (formal $\epsilon-\delta$) definition fails in such a case.

Define $f: \Bbb R \to \Bbb R$ by:

$f(x) = 0, x < 0$
$f(x) = 1, x \geq 0$

This is clearly defined on "all of the real numbers", and it "jumps" at 0. So let's see why our definition fails.

Suppose $\epsilon = \dfrac{1}{2}$. I will show that for this $\epsilon$, we cannot find any $\delta > 0$ that works.

For suppose we could. If so, we have for any $0 < |x - 0| < \delta$ (that is, any $x$ in the range: $(-\delta,0) \cup (0,\delta)$) that:

$|f(x) - L| < \dfrac{1}{2}$ (or whatever our limit $L$ might be).

Well, if $x = a > 0$, this means:

$|1 - L| < \dfrac{1}{2}$, which happens when $\dfrac{1}{2} < L < \dfrac{3}{2}$.

However, if $0 < a < \delta$, then $-\delta < -a < 0$, so that $0 < |-a - 0| < \delta$, as well. And for this value of $x$, we must have:

$|f(a) - L| < \dfrac{1}{2}$, that is:

$|-L| = |L| < \dfrac{1}{2}$, in other words: $-\dfrac{1}{2} < L < \dfrac{1}{2}$.

But a real number $L$ CANNOT BE greater than AND less than $\dfrac{1}{2}$ simultaneously, that's a contradiction.

So our assumption that we could find a $\delta$ that works cannot be correct, that is: there isn't any such $\delta$.

Since limits have to "go through" for ANY $\epsilon > 0$, this one counter-example is sufficient to show that:

$\lim\limits_{x \to 0} f(x)$ does not exist (in fact, choosing any $\epsilon < 1$ causes our limit test to fail, but the $\epsilon = \dfrac{1}{2}$ case is "easy to prove").

Intuitively (and sometimes formally), one typically sees this by noting the "right-hand limit" and the "left-hand limit" do not agree (by using $\epsilon =$ "half the jump distance").

Now, I know this involves (especially for more complicated functions) a lot of ugly algebra, but it's the way mathematicians show they are "smarter than the pencil".

 

[samir]

So the point needs to be an element in the domain, and the limit at that point has to be equal to the function value at that point? I say point, but I really mean x value.

$$\lim_{{x}\to{a}}f(x) = f(a)$$

So unless we have a graph of the function, we have to study the the limit at any given point that we suspect there might be a discontinuity?

There appear to be at least two different cases where a point discontinuity can occur. One is when a function is defined specifically for an isolated x-value, as with the piece-wise functions above. Or when the natural domain of the function rule dictates that the function is undefined at that point, like with rational functions.

 

[I like Serena]

So the point needs to be an element in the domain, and the limit at that point has to be equal to the function value at that point? I say point, but I really mean x value.

$$\lim_{{x}\to{a}}f(x) = f(a)$$

So unless we have a graph of the function, we have to study the the limit at any given point that we suspect there might be a discontinuity?

There appear to be at least two different cases where a point discontinuity can occur. One is when a function is defined specifically for an isolated x-value, as with the piece-wise functions above. Or when the natural domain of the function rule dictates that the function is undefined at that point, like with rational functions.

Hey samir! ;)

The word point is a bit ambiguous, since it may mean either a point like $(x,y)$ in the plane, or a value $x$ in some domain of a function (which can actually also be a point in a plane). Anyway, in our current context point means a value in the domain of a function.

As Opalg pointed out, if a function is not defined at a point, it is neither continuous nor discontinuous - since the concept of continuity itself is not defined at that point.

Discontinuities of real functions at a point are usually classified in the following categories:

1. Removable discontinuity (can be removed by redefining the function at the point).
2. Jump discontinuity (left and right limits both exist but are different).
3. Essential discontinuity (left and/or right limit does not exist).

There's also the edge case of an isolated point, but this is not a discontinuity.
A function at an isolated point is continuous.

Sorry, no symbols in this explanation.