A function 'continous' at a 'point'.

  • Context: Undergrad 
  • Thread starter Thread starter dE_logics
  • Start date Start date
  • Tags Tags
    Function Point
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
32 replies · 5K views
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.
 
Physics news on Phys.org
dE_logics said:
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)---- ------------ ------ ------
 
dE_logics said:
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 continuous 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 continuous 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.