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Don't understand rules of continuity/discontinuity with Limits

  1. May 22, 2014 #1
    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...: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.
     
  2. jcsd
  3. May 22, 2014 #2

    micromass

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    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:

    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.
     
  4. May 22, 2014 #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
     
  5. May 22, 2014 #4

    micromass

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    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

    [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##.
     
  6. May 22, 2014 #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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