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Thread: Proving differentiability

  1. #1
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
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    288

    Proving differentiability

    Let f(x)=
    x^2 if x rational
    0 if x is irrational
    Prove f is differentiable at 0

    f(0)=0
    So at $\displaystyle x=0$
    $\displaystyle lim_{h\rightarrow0}=\frac{f(0+h)-f(0)}{h}
    $
    $\displaystyle lim_{h\rightarrow0}=\frac{f(h)}{h}
    $
    Case 1: h rational, then
    $\displaystyle lim_{h\rightarrow0}=\frac{h^2}{h}=h=0
    $
    Case 2: h irrational
    $\displaystyle lim_{h\rightarrow0}=\frac{0}{h}=0
    $
    Hence
    $\displaystyle lim_{h\rightarrow0}=\frac{f(0+h)-f(0)}{h}
    $
    and f is differentiable at 0
    QED

    Am I expressing this proof correctly? Are all my steps rigorous?
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  2. #2
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
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    This isn't quite right - you can't just pick and choose which values of $\displaystyle h$ you attain. A cleaner way to do this is to use the Sandwich theorem, combined with taking the limit from the right and from the left. First, from the right, we have $\displaystyle h>0$. Then, $\displaystyle 0\leq f(h)\leq h^2$. Dividing by $\displaystyle h$ gives $\displaystyle 0\leq \frac{f(h)}{h}\leq h$. Taking the limit as $\displaystyle h\to 0^+$ shows that $\displaystyle \frac{f(h)}{h}\to 0$.

    I'll leave the left-hand approach to you. Good luck.
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