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Proving that f'(0) exists

  1. Sep 27, 2010 #1
    I wanted to see what kind of responses I would get regarding this problem:

    Let f : [tex]\Re[/tex][tex]\rightarrow[/tex][tex]\Re[/tex] be a continuous function that is differentiable for all nonzero x such that f '(x) exists. If f'(x) [tex]\rightarrow[/tex] L as x[tex]\rightarrow[/tex]0 exists, prove that f '(0) exists.
  2. jcsd
  3. Sep 27, 2010 #2
    Use the Mean Value Theorem
  4. Sep 28, 2010 #3


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    A few LaTeX tips: Put tex or itex tags around the whole formula instead of around each symbol, and use \mathbb R for the set of real numbers. Example: [itex]f:\mathbb R\rightarrow\mathbb R[/itex]. (Click the quote button to see what I did).
  5. Sep 28, 2010 #4


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    While derivatives are not necessarily continuous, they do satisfy the "intermediate value property" (if f'(a)< c< f'(b), then f'(d)= c for some d between a and b). You can show that using the mean value theorem JG89 suggests. From that it follows that if [itex]\displaytype\lim_{x\to a}f'(x)[/itex] exists, then so does f'(a) and [itex]\displaytype f'(a)= \lim_{x\to a}f'(x)[/itex]
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