设 函数 f(x)在x=2处可导,且f(2)的导数=1求: lim f(2+h)—f(2—h)/2h h→0
3个回答
lim f(2+h)—f(2—h)/2h =lim[ f(2+h)-f(2)/2h—(f(2—h)-f(2))/2h ]
=f'(2)/2+f'(2)/2=1
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