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Integral delta

  1. May 12, 2009 #1
    This isn't homework per se... It's a question from a book I'm self-studying from.

    If [tex]f[/tex] is continuous on [tex][a,b][/tex] and differentiable at a point [tex]c \in [a,b][/tex], show that, for some pair [tex]m,n \in \mathbb{N}[/tex],

    [tex]\left | \frac{f(x)-f(c)}{x-c}\right | \leq n[/tex] whenever [tex]0 \leq |x-c| \leq \frac{1}{m}[/tex]

    ------

    Since it's differentiable at c I know [tex]\lim_{x \to c}\frac{f(x)-f(c)}{x-c}[/tex] exists...

    And it's continuous so I know for whatever [tex]\epsilon > 0[/tex] I pick there's a [tex]\delta > 0[/tex] so that [tex]|f(x)-f(c)| < \epsilon[/tex] when [tex]|x-c| < \delta[/tex]

    So I guess 1/m could be delta? And then... ? Not sure how to round this one off
     
  2. jcsd
  3. May 12, 2009 #2
    Re: problem

    Couldn't you just conclude that:

    [tex]\left | \frac{f(x)-f(c)}{x-c}\right | \leq n \Leftrightarrow | f(x)-f(c) | \leq n | x - c | \leq \frac{n}{m}[/tex]

    and that you can always find m and n such that this holds as long as [tex]| f(x)-f(c) |[/tex] is finite which is ensured by differentiability?
     
  4. May 13, 2009 #3

    Shooting Star

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    Re: problem

    Pick m as the integral part of 1/δ so that 1/δ = m+f. Then δ = 1/(m+f) < 1/m.
     
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