设f(x)在x=3点可导,则lim{[f(3-h)-f(3)]/2h}=? h →0

设t=-h,h →0 ,则t →0 h →0 lim{[f(3-h)-f(3)]/2h}= t →0 lim{[f(3+t)-f(3)]/(-2t)}=(-1/2) t →0lim{[f(3+t)-f(3)]/t}f(x)在x=3点可导则按导数定义t →0 lim{[f(3+t)-f(3)]/t}=f'(3)所以答案为(-1/2)f'(3)...