# Don't understand rules of continuity/discontinuity with Limits

1. May 22, 2014

Hi,

My text book is giving me 3 rules that must be met to be continuous:

I don't understand what my text book is trying to explain to me,

- ƒ is defined at a...

-Lim ƒ(x) = ƒ(a)...
x→a

all i undersdand is that if you draw a graph that is continuous, you must be able to draw the graph without lifting your pen.

2. May 22, 2014

### micromass

Staff Emeritus
You only seem to be giving 2 rules below?

That characterization isn't entirely correct, but it's a good intuition.

It would help if you would explain what you don't understand about the rules you've given above.

A good thing to try is to invent some discontinuous functions yourself and to see how the rules fail. One of the most famous examples is the Heaviside step function:

So this function is defined as a function $H:\mathbb{R}\rightarrow \mathbb{R}$ such that $H(x) = 0$ if $x<0$, $H(x)=1$ if $x>0$ and $H(0) = 1/2$. You can see from the graph that the function is not continuous and that there is a problem in $0$. Does this function satisfy

$$\lim_{x\rightarrow 0} H(x) = H(0)$$

Here's another function:

Is this function continuous? Where is the problem? What rule doesn't it satisfy?

Another one:

This function to be discontinuous. The reason is of course that the function is undefined at $x=3$, so $f(3)$ is not defined. According to your definition, for the function to be continuous at $3$, it must be defined there.

3. May 22, 2014

Hi micromass,

The third rule is:

Lim ƒ(x) exists
x→a

I don't understand the statement: H:R→R

I do understand H(x) = 0 if x < 0, and
H(x) = 1 if x > 1, from looking at the graph.

So if something is undefined it is:

2/0 or ∞/0, is 0/0 undefined or is it just 0?

Here is one i tried to make up:

ƒ(x) = x + 3 (1)
x→a

then;

ƒ(1) = 1 + 3
= 4

would that exampe state the ƒ(x) ≠ ƒ(a)

because (1) = 4...discontinuous

4. May 22, 2014

### micromass

Staff Emeritus
OK, makes sense.

It means that the domain of $H$ is $\mathbb{R}$ and the codomain $\mathbb{R}$. So this means that $H$ is a function that takes in elements of $\mathbb{R}$ and outputs elements of $\mathbb{R}$. For example, you can input $-2$ and get as output $0$.

No, it's just undefined. It doesn't have any value. For example, if I define the function $F$such that $F(x) = 0$ for $x<0$ and $F(x) = 0$ for $x>0$. Then I haven't stated what $F(0)$ is. Thus $F$ is undefined in $0$.

The $x\rightarrow a$ should be under a limit. I take it that you define the function $f(x) = x+3$. This is fine.

Let's investigate continuity in $a=5$. On one hand, we have $f(5) = 5+3 = 8$. On the other hand, we can prove that

$$\lim_{x\rightarrow 5} f(x) = \lim_{x\rightarrow 5} x+3 = 8$$

So we see that

$$\lim_{x\rightarrow 5} f(x) = f(5)$$

so the function is continuous in $5$. You can do the same with all other values of $a$ in this case.

On the other hand, let's define $g(x) = x+3$ for each $x\neq 5$ and let's define $g(5) = 0$. This is also perfectly possible. Then we have $g(5) = 0$, but

$$\lim_{x\rightarrow 5} g(x) = \lim_{x\rightarrow 5} x+3 = 8$$

So we have that

$$\lim_{x\rightarrow 5} g(x) \neq g(5)$$

hence the function is not continuous in $5$.

5. May 22, 2014

### gopher_p

Just to add to what micromass has already said, "$f$ is defined at $a$" is essentially just another way of saying "$a$ is in the domain of $f$".

If you write the three conditions out as

(1) $\lim_{x\rightarrow a} f(x)$ exists
(2) $f$ is defined at $a$
(3) $\lim_{x\rightarrow a} f(x)=f(a)$

then conditions (1) and (2) ensure that it makes sense to talk about, respectively, the left and right hand sides of the equation in (3). For instance, if $f$ is not defined at $a$, then the term (i.e. collection of symbols) "$f(a)$" is literally meaningless, and the equation in (3) is nonsense. Or to put it another way, the three conditions could be stated in (slightly) more colloquial language as

(1) $\lim_{x\rightarrow a} f(x)$ is a number
(2) $f(a)$ is also a number
(3) $\lim_{x\rightarrow a} f(x)$ and $f(a)$ are the same number.