如何证明lim[f(x+h)+f(x-h)-2f(x)]=f"(x) 其中h趋向0

若f(x)有二阶导数,证明f''(x)=lim(h→0)f(x+h)-2f(x)+f(x-h)/h^2. f(x)在x处二阶可导,求lim{[f(x+h)-2f(x)+f(x-h)]/h^2},h趋向于0 设f(x)具有二阶导数f''(x),证明f''(x)=lim(f(x+h)-2f(x)+f(x-h))/h^2 f(x)在x=a处可导, lim(h→0) [f(a+h)-f(a-2h)]/h= 设函数f(x)在x=x0处可导,则lim(h>0)[f(x0)-f(x0-2h)]/h 设f(x)在x=a处可导,f'(x)=b 求极限lim(h-0) f(a-h)-f(a+2h)/ h 设f'(x) = 3^(1/2) ,求 lim(h→0) [f(x+mh) - f(x - nh)] / h ,(m , 设f(x)在x=2处可导,且f'(2)=1,则lim h→0 [ f(2+h)-f(2-h)]/h等于多少, 举例说明lim(h→0)f(xo+h)-f(xo-h)\2h=f'(xo)存在,推导不出函数f(x)在x=xo 设f(X)在x=x0处具有二阶导数f''(x0),试证:lim(h→0)(f(x0+h)-2f(x0)+f(x0-h)) 导数极限形式的证明1）f'(x0)=lim(x→x0)[f(x)-f(x0)]/(x-x0) 2)f'(x)=lim(h 正弦函数用定义求导就是用定义来求 F(x)=Sin(x)F'(x)=Lim h->0 ( (F(x+h)-F(x))/h