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Math Help - Differentiability of functions

  1. #1
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    Differentiability of functions

    If the function f is defined as follows:
    f(x)=x^2 for x rational
    f(x)=0 for x irrational
    How would you show that f is differentiable at x=0?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    If the function f is defined as follows:
    f(x)=x^2 for x rational
    f(x)=0 for x irrational
    How would you show that f is differentiable at x=0?
     f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

    We need to show this limit exists when  x=0 .

    For this we need to compare  \limsup and  \liminf .

    For  x\neq 0 ,  x^2 > 0 , so our  \limsup will only consider the sequence containing all points of the form  (x,x^2) .

    Thus  \limsup_{h\to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h\to 0}\frac{(0+h)^2-0^2}{h} = \lim_{h\to 0}\frac{h^2}{h} = \lim_{h\to 0}h = 0 .

    Clearly our  \liminf will only consider the sequence containing all points of the form  (x,0) .

    Thus  \liminf_{h\to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h\to 0}\frac{0}{h} = 0 .

    Since  \limsup=\liminf ,  \lim_{h\to 0}\frac{f(0+h)-f(0)}{h} exists and equals  0 .
    Therefore  f'(0)=0 .
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