This may be ridiculous but I am just starting to learn Calculus.

[Loismustdie]

This may be ridiculous but I am just starting to learn Calculus.
Anyway I am learning continuity and they say the function is continuous at a certain point c. What I want to know is what does this mean, and how do you know if the function is completely continuous.

 

[arildno]

For starters, have you bothered to read the definition of continuity?

 

[Loismustdie]

yeah but I am still confused about continuity. If you could answer my question maybe I would know

 

[arildno]

Vague confusions are not easily resolved!

Please pick up your book again, and state what SPECIFICALLY is unclear to you about the definition of continuity!

 

[Loismustdie]

My books definition states that "A function f(x) is continuous at x=c if lim(x->c)f(x)=f(c)." What I am confused about is that if it is continuous at the point x=c how do you know if the whole function is continuous and not just at x=c.

Help me if this can be explained, if not I will just drop it.

 

[matt grime]

The function is continuous at all points if the above holds for all c.

 

[arildno]

"What I am confused about is that if it is continuous at the point x=c how do you know if the whole function is continuous and not just at x=c.
"

In order to know that the function is continuous at ALL its points, then it has to fulfill the condition of continuity at a point everywhere.

 

[HallsofIvy]

My books definition states that "A function f(x) is continuous at x=c if lim(x->c)f(x)=f(c)." What I am confused about is that if it is continuous at the point x=c how do you know if the whole function is continuous and not just at x=c.

Help me if this can be explained, if not I will just drop it.

If all you know is that f is continuous at x= c, then you do not know anything about whether it is continuous for other x. For example, the function f(x)= 1/n if x= m/n reduced to lowest terms, f(x)= 0 if x is irrational is continuous at x= 0 but no where else. (I think!)

 

[arildno]

I don't agree, Halls!
Your function ought to be continuous on all irrationals, but discontinuous at the rationals.

Discontinuity at the rationals is trivial, here's how I think a proof of continuity at the irrationals might go:

1. Given an irrational a, for every N, there exists a D(N)>0 so that the punctured disk about a with radius D(N) do not contain any rationals of the form m/N.
2. Given E>0, choose 1/N<E
For n=1,...N, find the minimum radius D(n*).
3. Thus, only rationals with denominator N+1 or greater will be included in the indicated punctured disk, by whom the associated function values will be less than E.

Hence, continuity at the irrationals is proven.

 

[HallsofIvy]

Okay, arildno, I concede. I had a suspicion I was wrong when I wrote that!

 

[shimigoloo]

Hi
I think that you imagine continuty as a geometrical concept.thus you think about that the founctions`s graph is a continuous one or not.
In strict mathematical conception , a (whole) function is continuous if for every point c in it`s domain satisfy continouty condition.
( in the parantesis in the calculus there is atheoreom that is say : if f(x) is continuous in the interval [a,b] and f(a)=k1 and f(b)=k2; then for every k in which k1<k<k2 (or k2<k<k1) , there is a c in [a.b] in which f(c)= k)

 

[arildno]

Well, the idea that the graph of a conintuous function is something you can draw without lifting your pen is rather outmoded.

Furthermore, looking at nastinesses like the continuous, nowhere differentiable Weierstrass function drawing the graph becomes everywhere impossible!

Also, if the domain is not connected, then the graph of a continuous function looks not at all..connected.

 

[mathwonk]

a nice example is f(x) = x if x is rational, and f9x) = -x if x is irrational.

this function is con tinuous only at x=0, and is proabbly the exampel halls was thinking of.

try to draw the graph and you will get a better idea of what con tinuity means.

i.e. f(0) can be approximated by any f(x) for any x near enough to x=0, but this is not true anywhere else.