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A function 'continous' at a 'point'.

by dE_logics
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dE_logics
#19
Oct6-09, 10:23 PM
P: 735
No, cause I do not understand the notion of limit.

Can you please tell me in terms of ε and δ?
dE_logics
#20
Oct7-09, 10:32 PM
P: 735
HallsofIvy
#21
Oct8-09, 06:04 AM
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You keep saying you "do not know limits". How can you hope to understand, or even ask about, continuity, then?

But since you ask: "[itex]\lim_{x\to a} f(x)= L[/itex]" if and only if, given any [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]0< |x- a|< \delta[/itex], then [itex]|f(x)- L|< \epsilon[/itex].

That basically means, as just about every other response here has said, that you can f(x) as close to L as you like just by taking x close enough to a.

In order that a function, f(x), be continuous at x= a, three things must be true:
1) f(a) must exist.
2) [itex]\lim_{x\to a} f(x)[/itex] must exist.
3) [itex]\lim_{x\to a} f(x)[/itex] must be equal to f(a).

Since just writing [itex]\lim_{x\to a} f(x)= f(a)[/itex] pretty much implies that the two sides exist, most of the time we just write that.
statdad
#22
Oct8-09, 02:18 PM
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P: 1,377
You stated that [tex] f [/tex] is continuous at [tex] c [/tex] if, for every [tex] \epsilon > 0 [/tex], there is a [tex] \delta > 0 [/tex] such that

[tex]
| f(c+h) - f(c)| < \epsilon \quad \text{ if } |h| < \delta
[/tex]

An intuitive statement of continuity at a point is this:

The function [tex] f [/tex] is continuous at a number [tex] c [/tex] if [tex] f(c+h)[/tex] is
close in value to [tex] f(c) [/tex] when [tex] c+h [/tex] is close to [tex] c [/tex].
  • [tex] \epsilon [/tex] shows how close you want [tex] f(c+h) [/tex] to be to [tex] f(c) [/tex]
  • [tex] \delta [/tex] measures how close to [tex] c [/tex] you need to pick other x-values to make the previous point come true
  • [tex] h [/tex] measures the distance the new x-value is from the number [tex] c [/tex]
dE_logics
#23
Oct12-09, 09:59 AM
P: 735
Quote Quote by HallsofIvy
You keep saying you "do not know limits". How can you hope to understand, or even ask about, continuity, then?
That's cause in every book first they explain continuity, then limit.

Quote Quote by statdad View Post
You stated that [tex] f [/tex] is continuous at [tex] c [/tex] if, for every [tex] \epsilon > 0 [/tex], there is a [tex] \delta > 0 [/tex] such that

[tex]
| f(c+h) - f(c)| < \epsilon \quad \text{ if } |h| < \delta
[/tex]

An intuitive statement of continuity at a point is this:

The function [tex] f [/tex] is continuous at a number [tex] c [/tex] if [tex] f(c+h)[/tex] is
close in value to [tex] f(c) [/tex] when [tex] c+h [/tex] is close to [tex] c [/tex].
  • [tex] \epsilon [/tex] shows how close you want [tex] f(c+h) [/tex] to be to [tex] f(c) [/tex]
  • [tex] \delta [/tex] measures how close to [tex] c [/tex] you need to pick other x-values to make the previous point come true
  • [tex] h [/tex] measures the distance the new x-value is from the number [tex] c [/tex]
PERFECT answer man...just perfect...thanks.
HallsofIvy
#24
Oct12-09, 10:29 AM
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Quote Quote by dE_logics View Post
That's cause in every book first they explain continuity, then limit.
In every book? Peculiar- I have never seen such a book!



PERFECT answer man...just perfect...thanks.[/QUOTE]
Yes, it was.
dE_logics
#25
Oct13-09, 12:23 AM
P: 735
Yes, check out these books -

DIFFERENTIAL CALCULUS - SANTI NARAYAN
Introduction to calculus - KAZIMIERZ KURATOWSKI
Calculus - Benjamin Crowell

Actually there are no books where limits is discussed before continuity.

So the value of h has 2 restrictions...one ε and the other δ.

Is the value of δ arbitrator?
Office_Shredder
#26
Oct13-09, 12:43 AM
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The value of delta depends on the value of epsilon. The value of epsilon is arbitrary.

I'm looking at Introduction to calculus by Kuratowski right now. Section 4: Functions and their Limits

Section 5: Continuous functions

I notice that it defines limits in terms of sequences but these are equivalent. At any rate, the book definitely defines limits before continuity
dE_logics
#27
Oct13-09, 06:01 AM
P: 735
hummm...yes, you're right about Kuratowski.

Exclude that book and Benjamin Crowell has continuity at 2.8 and limits at 2.9
emyt
#28
Oct13-09, 07:22 AM
P: 218
Quote Quote by dE_logics View Post
Yes, check out these books -

DIFFERENTIAL CALCULUS - SANTI NARAYAN
Introduction to calculus - KAZIMIERZ KURATOWSKI
Calculus - Benjamin Crowell

Actually there are no books where limits is discussed before continuity.

So the value of h has 2 restrictions...one ε and the other δ.

Is the value of δ arbitrator?
Calculus - Michael Spivak
Introduction to Calculus and Real Analysis - Richard Courant and John Fritz

both these books are popular, and they have a treatment on limits before continuity
dE_logics
#29
Oct13-09, 10:25 AM
P: 735
I do have Introduction to Calculus and Real Analysis - Richard Courant and John Fritz somewhere...anyway, I'll get them.

Thanks.
dE_logics
#30
Oct13-09, 10:53 PM
P: 735
This is the verdict; hope it helps someone -

Consider an example...lets have a function such that for an interval a to b the function returns values continuously but below a or below b it returns one values (for the respective limit cross (below a and above b)).
First we will like to have the list of variables -
ε – An 'arbitrary' value which is supposed to be related to f(x)......in some way...after including a variable 'h'.
x – X server...just kidding. The independent variable of the function f.
δ – 'h' is made to assume this value actually such that |h|<δ; δ is also dependent on ε and it's value is specific. The value of h can be varied arbitrarily, but it has a restriction |h|<δ...this is the limit of h's arbitrary value.
h – A value added/subtracted to x; since it will be added/subtracted to x, f(x + |h|) or f(x – |h|), it will return a deviated value as compared to f(x). It's a criteria that the deviation should be < ε...in mathematical terms |f(x+|h|)-f(x)| < ε or |f(x-|h|)-f(x)| < ε; notice that|h|<δ, the value of δ is such that it will not permit |f(x+h)-f(x)| or |f(x-|h|)-f(x)| to be greater and equal to ε...or the value of |h| should not exceed δ. So the value of δ is based on the restriction posed by ε and |h| can be varied by |h|<δ.
A function is said to be continuous iff for a very small value of ε the corresponding value of δ is also small and cause value of δ will be small the rigidity in value of h will too be high; notice that by this definition, even if |f(x+|h|)-f(x)| = 0, the function will be considered continuous.
If taking a fresh example of the function g(e) = y; for an infinitely small change in value of e, there should be an infinitely small change or no change in value of y for the function to be continuous.
dE_logics
#31
Oct14-09, 03:12 AM
P: 735
aaaa....but still there is a major doubt left.

What do you mean by a 'point'...for a function to be continuous at a 'point', the value of ε should be assumed infinitely small and the corresponding value of δ should also be infinitely small...THEN we can call the function continuous at a 'point'...and that point will be f(x)...or the value of f at x.
emyt
#32
Oct21-09, 11:30 AM
P: 218
Quote Quote by dE_logics View Post
aaaa....but still there is a major doubt left.

What do you mean by a 'point'...for a function to be continuous at a 'point', the value of ε should be assumed infinitely small and the corresponding value of δ should also be infinitely small...THEN we can call the function continuous at a 'point'...and that point will be f(x)...or the value of f at x.
if a function is continuous at a point, then there will exist some kind of neighbourhood around that point such that f(a+h) as h approaches 0 is f(a).

-------xxxxxxx--f(a)--f(a+h)--xxxxxx------xxxxx -------

imagine this line as a discontinuous function (the x's are spaces, spaces don't work here), the dashes can be infinitesimally small distances, but the function is continuous at a because there is a neighbourhood around it where you can pick f(a+h) and have the limit of that as h goes to zero as f(a)


----- xxxxx-- xxx-------xxxxxxxf(a)xxxxxxx---- xxxxx--------- -----xxxxx ------


if f(a) is defined like this, then there is no neighbourhood around f(a) such that f(a+h) as h approaches 0 is f(a)


----- --- ------- xxxxxxf(a)xxxxxxxx----f(a+h)-xxxxxxxx------------ ------ ------



----- --- ------- xxxxxxxx f(a)xxxxxxxx--f(a+h closer..)---xxxxxxxx ------------ ------ ------


----- --- -------xxxxxxxxxx f(a)xxxxxxxxxx-f(a+h cannot go any closer)---- ------------ ------ ------
HallsofIvy
#33
Oct22-09, 06:27 AM
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Quote Quote by dE_logics View Post
aaaa....but still there is a major doubt left.

What do you mean by a 'point'...for a function to be continuous at a 'point', the value of ε should be assumed infinitely small and the corresponding value of δ should also be infinitely small...THEN we can call the function continuous at a 'point'...and that point will be f(x)...or the value of f at x.
There is no such thing as "infinitely small" real numbers. You can do calculus in terms of "infinitesmals" but that requires extending the real numbers to a new number system and that is very deep mathematics. Certainly nothing you have said so far implies that you are familiar with infinitesmals and I recommend avoiding them in favor of the "limit" concept we have been using so far.

Saying that a function is "continuous at a point", say "f(x) is continuous at x= a", is exactly what we have been talking about here. "f(x) is continous at x= a" if and only if
1) f(a) exists.
2) [itex]\lim_{x\to a}f(x)[/itex] exists.
3) [itex]\lim_{x\to a} f(x)= f(a)[/itex].

More fundamentally, including the definition of "limit" in that definition
"Given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex] then [itex]|f(x)- f(a)|< \epsilon[/itex]".

The usual definition of "continuous" is "continuous at a point". We then extend the concept by saying that f(x) is "continous on a set" if and only if it is continous at every point of that set.

Saying that a function is continuous "at a point" does not restrict the possible values of [itex]\delta[/itex] and [itex]\epsilon[/itex] in any way.


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