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Homework Help: Why does differentiability imply continuity?

  1. Dec 26, 2007 #1
    I've been thinking... Since derivatives can be written as:

    [tex]f'(c)= \lim_{x\rightarrow{c}}\frac{f(x)-f(c)}{x-c}[/tex]

    and for the limit to exist, it's one sided limits must exist also right?

    So if the one sided limits exist, and thus the limit as x approaches c (therefore the derivative at c) (but f(x) is not continuous at c) why can't f(x) have a derivative at c?

    I'm just looking at it from that standpoint (I know that derivatives are basically the rate of change of a function at a point or in general).
  2. jcsd
  3. Dec 26, 2007 #2


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    http://people.hofstra.edu/stefan_Waner/RealWorld/calctopic1/contanddiffb.html [Broken]
    Last edited by a moderator: May 3, 2017
  4. Dec 26, 2007 #3


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    If the limit as x->c of f(x) does not exist and equal f(c) then the numerator doesn't tend to zero and the denominator does. How can the derivative exist? If you are asking if a function can have a one sided derivative, the answer is that it can.
    Last edited: Dec 26, 2007
  5. Dec 27, 2007 #4
    Derivatives are only defined for points in the domain of a function

    if c is not in the domain of f(x), then f'(c) by definition does not exist
  6. Dec 27, 2007 #5


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    True, but what does that have to do with continuity which was the whole point of the question?
  7. Dec 27, 2007 #6
    I think what chicken_dude wants to say is that to differentiate a function it has to be continuous already. Hence diff. implies continuity. But the reverse is not true.
  8. Dec 27, 2007 #7


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    Yes, but what does that have to do with "not being in the domain of the function"? I agree that if a point is not in the domain of a function, the function cannot be either continous or differentiable there, but the real question is about why, if a function is differentiable at a point in the domain the function must be continuous there.

    The function f(x)= x for x< 1, x+ 1 for [itex]x\ge 1[/itex] has x= 1 in its domain but is neither continous nor differentiable there. Your initial reponse implied that you thought that "f is not continous at p" was the same as "p is not in the domain of f" which is certainly not true.
    Last edited by a moderator: Dec 27, 2007
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