Don't understand rules of continuity/discontinuity with Limits

In summary: And the point of the examples with the Heaviside step function and the others is that they show that the three conditions really are necessary for continuity. In a sense, the point of doing mathematics is that we want to be able to make rigorous statements that are true, and the three conditions are just a way of making a precise statement of what true means in this context.
  • #1
HadanIdea
6
0
Hi,

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

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

- ƒ is defined at a...:confused:

-Lim ƒ(x) = ƒ(a)...:confused:
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.
 
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  • #2
HadanIdea said:
Hi,

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

You only seem to be giving 2 rules below?

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

- ƒ is defined at a...:confused:

-Lim ƒ(x) = ƒ(a)...:confused:
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.

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:

325px-Dirac_distribution_CDF.svg.png


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

[tex]\lim_{x\rightarrow 0} H(x) = H(0)[/tex]

Here's another function:

continuous_4.gif


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

Another one:

continuous_5.gif


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
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
HadanIdea said:
Hi micromass,

The third rule is:

Lim ƒ(x) exists
x→a

OK, makes sense.

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

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##.

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?

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##.

Here is one i tried to make up:

ƒ(x) = x + 3 (1)
x→a

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

[tex]\lim_{x\rightarrow 5} f(x) = \lim_{x\rightarrow 5} x+3 = 8[/tex]

So we see that

[tex]\lim_{x\rightarrow 5} f(x) = f(5)[/tex]

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

[tex]\lim_{x\rightarrow 5} g(x) = \lim_{x\rightarrow 5} x+3 = 8[/tex]

So we have that

[tex]\lim_{x\rightarrow 5} g(x) \neq g(5)[/tex]

hence the function is not continuous in ##5##.
 
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  • #5
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.
 
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1. What is continuity and discontinuity in terms of limits?

Continuity in terms of limits refers to the smoothness of a function at a particular point. A function is continuous at a point if the limit of the function at that point exists and is equal to the value of the function at that point. On the other hand, discontinuity occurs when the limit of a function at a point does not exist or is not equal to the value of the function at that point.

2. How do you determine if a function is continuous or discontinuous at a point?

A function is continuous at a point if the limit of the function at that point exists and is equal to the value of the function at that point. To determine this, you can use the continuity criteria, which state that for a function to be continuous at a point, the left-hand limit and the right-hand limit must both exist and be equal to the value of the function at that point.

3. What types of discontinuity can occur in a function?

There are three main types of discontinuity: removable, jump, and infinite. A removable discontinuity occurs when there is a hole in the graph of the function at a particular point. A jump discontinuity occurs when the left and right-hand limits at a point exist, but they are not equal. An infinite discontinuity occurs when the limit of the function at a point is either positive or negative infinity.

4. Can a function be continuous at some points and discontinuous at others?

Yes, a function can be continuous at some points and discontinuous at others. This is because continuity is determined at a point-by-point basis, and a function can have different behaviors at different points. For example, a function can be continuous at all points except for one where there is a jump or infinite discontinuity.

5. How do limits relate to continuity and discontinuity?

Limits play a crucial role in determining continuity and discontinuity of a function. As mentioned before, a function is continuous at a point if the limit of the function at that point exists and is equal to the value of the function at that point. On the other hand, if the limit does not exist or is not equal to the value of the function at that point, then the function is discontinuous at that point. Therefore, limits help us determine the behavior of a function at a particular point and whether it is continuous or discontinuous.

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