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Thread: Differentiability implies continuity question

  1. #1
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    Question Differentiability implies continuity question

    Hi,

    So I'm trying to fully understand the theorem that suggests if "f" is differentiable at "a", then "f" is continuous at "a".

    The issue that i'm facing with this definition is when I'm dealing with function which have holes in them, such as f(x) = (x^2-4)/(x-2) which then simplify's to f(x) = x+2, such that x does not equal 2.

    So even though f(x) is differentiable, the function is not continuous given that it has a hole at x = 2. Can someone please let me know whether or not my logic is right about this? This is the only idea that is getting in the way of me understanding the relationship between differentiability and continuity. Really what I'm doing here is disproving the theorem that I stated above (I believe), and I would really appreciate someone to explain why or why not I may be wrong.

    Please help!
    Olivia
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    Re: Differentiability implies continuity question

    The function is not differentiable at $x=2$ either (the function does not exist to be able to take the derivative). The definition of the derivative at $x=2$ is:

    $\displaystyle \lim_{x\to 2} \dfrac{f(x)-f(2)}{x-2}$

    But $f(2)$ does not exist, so the limit does not exist.
    Last edited by SlipEternal; Jul 20th 2017 at 11:11 PM.
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    Re: Differentiability implies continuity question

    Got it, thanks!
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    Re: Differentiability implies continuity question

    Quote Originally Posted by otownsend View Post
    if "f" is differentiable at "a", then "f" is continuous at "a".
    This question has nagged at me for some. Because I read the rest of the post as well as the reply.

    I went to my library to find a textbook that I had written a pre-publication review. I found it is a box that had not looked at for twenty-five years. The text is Foundation of Analysis by Belding & Mitchell.
    Their definition of continuity is: a function $f$ is continuous as $x=a$ provided $\large\displaystyle\lim_{h\to 0}f(a+h)=f(a)$. I remember think that is a clever reduction. Then several sections later it was applied brilliantly to your question.

    Suppose that the function $f$ is such that $f'(a)$ exists then the function $f$ is continuous at $x=a$.

    $\displaystyle \lim_{h\to 0}f(a+h)-f(a)=$
    $\displaystyle \lim_{h\to 0}\left(\dfrac{f(a+h)-f(a)}{h}\cdot h\right) =$
    $\displaystyle \lim_{h\to 0}\left(\dfrac{f(a+h)-f(a)}{h}\right)\cdot \lim_{h\to 0} h =f'(a)\cdot 0 =0$

    So we have $\displaystyle \lim_{h \to 0}f(a+h)-f(a)=0$ which is the definition of continuity.
    Last edited by skeeter; Jul 22nd 2017 at 06:55 PM.
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    Re: Differentiability implies continuity question

    Plato ... tried to fix the Latex ... see if I screwed it up. Sorry.
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