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B Continuous but Not Differentiable

  1. Mar 30, 2016 #1
    Suppose a certain function in continuous at [itex]c[/itex] and [itex](c. f(c))[/itex] exists, then which of the two could be false: [itex]\displaystyle \lim_{x \rightarrow c^-} {f(x)} = \lim_{x \rightarrow c^+} {f(x)}[/itex], and [itex]\displaystyle f'(c)[/itex]?

    I feel like both could be false, because if the formal derivative at a point exists, then the left and right hand limits much be equal -- but the function could be [itex]f(x) = |x|[/itex], which means that [itex]\displaystyle \lim_{x \rightarrow 0^-} {f(x)} \neq \lim_{x \rightarrow 0^+} {f(x)}[/itex] (but [itex]\displaystyle \lim_{x \rightarrow 0} {f(x)} = 0[/itex]) and [itex]f'(0) = \frac {d}{dx}|0|[/itex] does not exist. I feel like I'm missing something, cause the nuances around continuity and differentiability have always been confusing (and vague) to me.

    Thank-you
     
  2. jcsd
  3. Mar 30, 2016 #2

    Samy_A

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    If f is continuous in c, then ##\displaystyle \lim_{x \rightarrow c^-} {f(x)} = \lim_{x \rightarrow c^+} {f(x)} =f(c)##.

    For ##f(x)=|x|##, what you have is that ##\displaystyle \lim_{x \rightarrow 0^-} \frac{f(x)}{x}=\lim_{x \rightarrow 0^-} \frac{-x}{x}=-1## and ##\displaystyle \lim_{x \rightarrow 0^+} \frac{f(x)}{x}=\lim_{x \rightarrow 0^+} \frac{x}{x}=1##. That's why the function has no derivative in 0.
     
  4. Mar 30, 2016 #3

    HallsofIvy

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    No, this is wrong. Both [itex]\displaystyle \lim_{x \rightarrow 0^-} {f(x)}[/itex] and [itex]\lim_{x \rightarrow 0^+} {f(x)}[/itex] exist and are equal to 0.
    If [itex]\lim_{x\to a} f(x)[/itex] exists then it must be true that [itex]\lim_{x\to a^-} f(x)[/itex] and [itex]\lim_{x\to a+} f(x)[/itex] exist and are equal. Perhaps you are thinking of the fact that the limits of the derivatives are not equal: [itex]\displaystyle \lim_{x \rightarrow 0^-} {f'(x)} \neq \lim_{x \rightarrow 0^+} {f'(x)}[/itex].
    (Note f'(x), not f(x).)

     
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