※ 引述《hongo1120 (hongo1120)》之銘言：
: f(x)= x, x rational
: 0, x irrational
: g(x)= x^2, x rational
: 0, x irrational
: show that (1) f is not differentiable at x=0
: (2) g is differentiable at x=0 and give g'(0)
: thanks
如推文照定義
f'(0)= lim (f(h)-f(0))/h= lim f(h)/h
h→0 h→0
h若為有理數--> f'(0)=1
若為無理數--> f'(0)=0, 所以顯然h-->0時這極限不存在
因此f(x)在0處不可微
g'(0)=lim (g(h)-g(0))/h= lim g(h)/h
h→0 h→0
h若為有理數--> g'(0)=lim h^2/h=0
h→0
若為無理數--> g'(0)=0,此極限存在
因此g(x)在0處可微,且值為0
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