f'(x)=2.则lim[f(x-h)-f(x+2h)]/2h
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f'(x)=2.则lim[f(x-h)-f(x+2h)]/2h
f'(x)=2.则lim[f(x-h)-f(x+2h)]/2h
f'(x)=2.则lim[f(x-h)-f(x+2h)]/2h
lim[f(x-h)-f(x+2h)]/2h
h→0
=lim[f(x-h)-f(x)+f(x)-f(x+2h)]/2h
h→0
=lim{[f(x-h)-f(x)]+[f(x)-f(x+2h)]}/2h
h→0
=lim[f(x-h)-f(x)]/2h+lim[f(x)-f(x+2h)]/2h
h→0 h→0
=lim[f(x-h)-f(x)]/[(-h)▪(-2)]+lim[f(x)-f(x+2h)/(-2h)▪(-1)]
h→0 h→0
=f′(x)/(-2)+f′(x)/(-1)
=-3f′(x)/2
=-3/2·2
=-3
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