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On differentiability on endpoints of an open interval

  1. Sep 5, 2014 #1
    Before asking a question I would first like to mention the definitions of limit of function and differentiality at x=p

    1) Limit of function (f) at x=p
    Let E be domain of f and p be a limit point of E. Let Y be the range of f.

    If there exists q∈E such that for all ε>0 there exists δ>0 such that for all x∈E for which d(x,p)<δ implies d(f(x),q)<ε. Then we say that f(x)->q as x->p or lim(x->p) f(x)=q

    2) Differentiable at x=p (a ≤ x ≤ b)
    Let E=[a,b] be domain of f. Let Y be the range of f.

    If there exists limit of lim(x->p) s(x) (s(x)=[f(x)-f(p)]/[x-p] , where a< x < b), then f is differentiable at x=p.

    Sorry for the long definition! This is my question.

    Let E=(a,b) be deomain of f. Let Y be the range of f. Isn't it differentiable at x=a?



    As you can see in the picture, we can find a limit 'q' of all ε>0 there exists δ>0 such that for all x∈E for which d(x,p)<δ implies d(f(x),q)<ε. Thus lim(x->a) s(x)=q (by definition 1) )

    Please do ask me if anything's unclear!
     

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  3. Sep 5, 2014 #2

    HallsofIvy

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    If we "Let E=(a,b) be domain of f" then f is not even defined at x= a or x= b so certainly not differentiable!
     
  4. Sep 5, 2014 #3
    but look how differentiability at x=p is defined : lim(x->p) s(x)
    and this is why i have written definition of limit of function
    because there is a close relationship between them

    let's look at an example
    f(x)=x^2 where domain(E) is E=(0,1) ∪ (1,2) and E⊂R (real number, metric space)
    because 1∉E, does this mean lim(x->1) f(x) does not exist?
    In fact it does exist, which is 1!
     
    Last edited: Sep 5, 2014
  5. Sep 5, 2014 #4

    HallsofIvy

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    In your first post, you defined s(x)= (f(x)- f(p))/(x- p). If f(p) is not defined, then neither is s(x) for any x.
     
  6. Sep 5, 2014 #5
    i still don't get it
    by my example, like you said at x=1 it is not defined but there is still a limit value lim(x->1)f(x)
    likewise, at x=a it is not defined but there can be a limit value, which is lim(x->a)s(x)
     
    Last edited: Sep 5, 2014
  7. Sep 5, 2014 #6

    statdad

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    Having a limit value at a point does not mean it is defined at that point. Your difference quotient would be
    [tex]
    \dfrac{f(x) - f(1)}{x-1}
    [/tex]

    If [itex] f(1) [/itex] is not defined, this quotient is not defined.
     
  8. Sep 5, 2014 #7
    E=(0,1) union (1,2)
    f is a function from E to R
    s is a function from E to R

    f(x)=x^2
    s(x)=[f(x)-f(1)]/[x-1]

    hmm then why is there a difference between existence of limit between these two even though f(x) or s(x) is not defined at x=1?
    lim(x->1) f(x)=1 (exists)
    lim(x->1) s(x) ? (doesn't exist?)
     
  9. Sep 5, 2014 #8

    statdad

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    The limit for s does not exist because f(1) is not defined, so there is no number to use in that spot in the numerator.
    It is true that your f does not have to be defined at x=1 to have a limit there, because the limit depends only on values of x that are 'close to, but not equal to' 1.
    Your fraction doesn't have that property: it requires a value be known for f(1): since f is not defined there it doesn't have a value, so s is not defined.
    You can't confuse a limit value with a function value, which is what you're doing.
     
  10. Sep 5, 2014 #9
    thanks a bunch :)
     
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