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Thread: Differentiability of functions

  1. #1
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    Differentiability of functions

    If the function f is defined as follows:
    $\displaystyle f(x)=x^2$ for x rational
    $\displaystyle f(x)=0$ for x irrational
    How would you show that f is differentiable at $\displaystyle x=0$?
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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:
    $\displaystyle f(x)=x^2$ for x rational
    $\displaystyle f(x)=0$ for x irrational
    How would you show that f is differentiable at $\displaystyle x=0$?
    $\displaystyle f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} $

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

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

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

    Thus $\displaystyle \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 $\displaystyle \liminf $ will only consider the sequence containing all points of the form $\displaystyle (x,0) $.

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

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