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Show that this function is differentiable

  1. Aug 12, 2017 #1
    1. The problem statement, all variables and given/known data


    2. The attempt at a solution
    I'm not really sure where to start. We just want to show that ##\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = 0##. I see that ##\lim_{x \to c} (x - c)^2 = 0##. I feel that this may be a simple trick of inequalities, but I am having a complete brain fart at the moment. Can anyone provide any direction? Thanks in advance for any response.
     
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  3. Aug 12, 2017 #2

    Krylov

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    What is ##f(c)## equal to? Can you then find an upper bound for ##\frac{|f(x) - f(c)|}{|x - c|}##?
     
    Last edited: Aug 12, 2017
  4. Aug 12, 2017 #3
    Aha, yes! I didn't even consider finding what ##f(c)## equals. Having ##|f(x)| \leq (x - c)^2## for all ##x \in I## forces ##f(c) = 0##. Now, to find ##f'(c)## we must now find $$\lim_{x \to c} \frac{f(x)}{x - c}$$, which we can show to be ##0##, if we can alternatively show that $$\lim_{x \to c}|\frac{f(x)}{x-c}| = \lim_{x \to c}\frac{|f(x)|}{|x-c|} = 0$$. Now, ##|f(x)| \leq (x - c)^2## for all ##x \in I##, also implies that for ##x \in I-\{c\}##, ##\frac{|f(x)|}{|x-c|} \leq |x - c|##. Since the latter's limit is 0, and both are point-wise positive, we have that $$\lim_{x \to c}|\frac{f(x)}{x-c}| = 0$$, and we are done.
     
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