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Differentiable implies continuous

  1. Aug 19, 2010 #1
    ehhh
     
    Last edited: Aug 19, 2010
  2. jcsd
  3. Aug 19, 2010 #2

    Mark44

    Staff: Mentor

    Is this a homework problem?

    This is really hard to read, with everything on one line, and the malformed fraction near the beginning. To make it more readable, use a pair of tex and /tex tags for each line.
     
  4. Aug 19, 2010 #3
    It wasn't responding to any of the changes I was making to the Latex code and I gave up. I'm not sure what was wrong. And it wasn't a homework problem. I've never taken analysis. I was just wondering if this proof works after reading the theorem.
     
  5. Aug 19, 2010 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Often you will find that you need to click on your web-reader's "refresh" button when you edit LaTex. Why it doesn't refresh when you save the edit, I don't know.

    I cannot, of course, see what you have deleted so I cannot directly answer your question (which apparently is about a specific proof that "differentiable implies continuous") but I can say this:
    f(x) is differentiable at x= a if and only if
    [tex]\lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}[/tex]
    exists. Since the denominator necessarily goes to 0, in order that the limit exist, the numerator must also go to 0: we must have
    [tex]\lim_{h\to 0} f(a+h)- f(a)= 0[/tex]
    If we let a+h= x then as h goes to 0, x goes to a so that is the same as
    [tex]\lim_{x\to a}f(x)- f(a)= 0[/tex]

    which is the same as
    [tex]\lim_{x\to a} f(x)= f(a)[/tex]
    For that to be true, we must have
    1) [itex]\lim_{x\to a} f(x)[/itex] exist
    2) f(a) exist
    3) the two are equal
    which is precisely the definition of "continuous at x= a".
     
  6. Aug 19, 2010 #5
    [tex]
    f'(c) = \lim_{x\to c}\frac{f(x) - f(c)}{x - c}
    [/tex] ...since it is differentiable at any arbitrary point.

    [tex]
    f'(c) = \frac{\lim_{x\to c}f(x) - \lim_{x\to c}f(c)}{\lim_{x\to c}(x - c)}
    [/tex] ...using properties of the limit (i think).

    [tex]
    f'(c) * \lim_{x\to c}(x - c) = f'(c) * 0 = 0 = \lim_{x\to c}f(x) - \lim_{x\to c}f(c)
    [/tex] ...the limit of x - c is 0. The limit of f(c) is f(c) so this implies that

    [tex]
    \lim_{x\to c}f(x) = f(c)
    [/tex] which proves continuity.

    I know this can't be right. I am just trying to learn some of this. Also, would a more rigorous proof than the one you posted be at all better? I've seen some very confusing ones with deltas and epsilons and wonder if this is overkill?
     
    Last edited: Aug 19, 2010
  7. Aug 19, 2010 #6
    This does not work, as the limit of the denominator is 0. Distributing the limit over a quotient is only valid if the limit exists and the limit of the denominator is non-zero. However, you can get a working proof if you multiply both sides by [itex]\lim_{x\rightarrow c} x - c[/itex].
    Going down to the level of deltas and epsilons is not necessary. There is a proof using only what you know about the definition of the derivative at a point, and the definition of continuous at a point, without explicitly using epsilon-delta form.
     
    Last edited: Aug 19, 2010
  8. Aug 20, 2010 #7
    OK, I think I see that. After that you can pull out the limit sign and cancel out the x - c and are left with [tex] 0 = \lim_{x\to c}f(x) - f(c)[/tex]. The proof is basically done.
     
    Last edited by a moderator: Aug 21, 2010
  9. Aug 20, 2010 #8
    Yep. Good job.
     
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