设f(x)在x=x0的邻近有连续的二阶导数，证明;limh→0f(x0+h)+f(x0-h)-2f(x...
答：lim(h->0) [f(x0+h)+f(x0-h)-2f(x0)]/h^2 (0/0) =lim(h->0) [f'(x0+h)-f'(x0-h)]/(2h) (0/0) =lim(h->0) [f''(x0+h)+f''(x0-h)]/2 =f''(x0)
如下图片：
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