
#19
Oct609, 10:23 PM

P: 730

No, cause I do not understand the notion of limit.
Can you please tell me in terms of ε and δ? 



#20
Oct709, 10:32 PM

P: 730





#21
Oct809, 06:04 AM

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PF Gold
P: 38,904

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. 



#22
Oct809, 02:18 PM

HW Helper
P: 1,344

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].




#23
Oct1209, 09:59 AM

P: 730





#24
Oct1209, 10:29 AM

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PERFECT answer man...just perfect...thanks.[/QUOTE] Yes, it was. 



#25
Oct1309, 12:23 AM

P: 730

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? 



#26
Oct1309, 12:43 AM

Mentor
P: 4,499

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 



#27
Oct1309, 06:01 AM

P: 730

hummm...yes, you're right about Kuratowski.
Exclude that book and Benjamin Crowell has continuity at 2.8 and limits at 2.9 



#28
Oct1309, 07:22 AM

P: 218

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 



#29
Oct1309, 10:25 AM

P: 730

I do have Introduction to Calculus and Real Analysis  Richard Courant and John Fritz somewhere...anyway, I'll get them.
Thanks. 



#30
Oct1309, 10:53 PM

P: 730

This is the verdict; hope it helps someone 




#31
Oct1409, 03:12 AM

P: 730

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. 



#32
Oct2109, 11:30 AM

P: 218

xxxxxxxf(a)f(a+h)xxxxxxxxxxx  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 xxxxxxxxxxf(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)xxxxxxxxf(a+h)xxxxxxxx      xxxxxxxx f(a)xxxxxxxxf(a+h closer..)xxxxxxxx      xxxxxxxxxx f(a)xxxxxxxxxxf(a+h cannot go any closer)    



#33
Oct2209, 06:27 AM

Math
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P: 38,904

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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