lim(x->0+)f(x) = lim x^2 sin1/x = 0(无穷小*有界 = 无穷小)
所以f(x)在0连续,
lim(x->0-)'(x) =lim 2xsin1/x+x^2cos1/x*-1/x^2=lim 2xsin1/x-cos1/x = 0-lim cos1/x
导数不确定
lim(x->0-)f'(x) =lim 2x =0
所以导数不存在
f'(x)=2xsin1/x-cos1/x x>0
2x x
设f（x）=x^2sin1/x x>0 x^2 x≤0
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