利用级数展开到二阶即可
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
2个回答
相关问题
-
f(x)在X0处二阶可导,证lim(h->0)[ f(x-h0)+f(x0+h)-2f(x0)]/h^2=f``(x0)
-
f(x)=x^2 求lim{f(x+h)-f(x)}/h h趋向于0
-
f(x)在x_0处可导,求lim h→0 f(x_0+h)-f(x_0-h)/5h 的值
-
设函数f(x)在x=x0处可导,则lim(h>0)[f(x0)-f(x0-2h)]/h
-
设函数f(x)在点x0处可导,求lim(h→0)(f(x0+h)-f(x0-h))/2h的值
-
设函数f(x)在点x0处可导,求lim(h→0)(f(x0+h)-f(x0-h))/2h的值
-
设函数f(x)在点x0处可导,求lim(h→0)(f(x0+h)-f(x0-h))/2h的值
-
设 函数 f(x)在x=2处可导,且f(2)的导数=1求: lim f(2+h)—f(2—h)/2h h→0
-
f'(x)=-2,求limf(x+h)-f(x-h)/h(h趋向于0)
-
设函数f(x)在x=1处可导,且f'(1)=2,则[lim(h→0)f(1-h)-f(1)]/h等于