问一个高等数学的问题...设f''(x)存在,求证 lim(h→0) [f(x+2h)-2f(x+h)+f(x)]/h^2=f''(x)
2019-03-29
问一个高等数学的问题...
设f''(x)存在,求证 lim(h→0) [f(x+2h)-2f(x+h)+f(x)]/h^2=f''(x)
优质解答
使用一次洛必达法则,再使用导数的定义
lim(h→0) [f(x+2h)－2f(x+h)＋f(x)]/h^2
＝lim(h→0) [2f'(x+2h)－2f'(x+h)]/(2h)
＝lim(h→0) [f'(x+2h)－f'(x+h)]/h
＝lim(h→0) ｛2×[f'(x+2h)－f'(x)]/(2h)－[f'(x+h)－f'(x)]/h｝
＝2×lim(h→0)[f'(x+2h)－f'(x)]/(2h)－lim(h→0)[f'(x+h)－f'(x)]/h
＝2×f''(x)－f''(x)
＝f''(x)
使用一次洛必达法则,再使用导数的定义
lim(h→0) [f(x+2h)－2f(x+h)＋f(x)]/h^2
＝lim(h→0) [2f'(x+2h)－2f'(x+h)]/(2h)
＝lim(h→0) [f'(x+2h)－f'(x+h)]/h
＝lim(h→0) ｛2×[f'(x+2h)－f'(x)]/(2h)－[f'(x+h)－f'(x)]/h｝
＝2×lim(h→0)[f'(x+2h)－f'(x)]/(2h)－lim(h→0)[f'(x+h)－f'(x)]/h
＝2×f''(x)－f''(x)
＝f''(x)