其请问 lim(h→0) [ f(x0+3h)-f(x0-2h) ] / h

拆成两部分
[ f(x0+3h)-f(x0-2h) ] / h=3*[f(x0+3h)-f(x0)]/3h+2*[f(x0-2h)-f(x0)]/(-2h)
于是根据极限的定义，h趋于0时，上式趋于3*f'(x0)+2*f'(x0)=5f'(x0)

lim(h→0) [ f(x0+3h)-f(x0-2h) ] / h
=5lim(h→0) [ f(x0+3h)-f(x0-2h) ] /[（x0+3h)-(x0-2h)]
=5f'(x0)

令x1=x0-2h（h→0时x1→x0) 则x0+3h=x1+5h
lim(h→0) [ f(x0+3h)-f(x0-2h) ] / h
=5*lim(h→0)[f(x1+5h)-f(x1)]/5h
=5f`(x1)=5f`(x0)