(B.)lim_(h to 0)(f(a+2h)-f(a+h))/(h)存在. (C.)lim_(h to 0)(f(a+h)-f(a-h))/(2h)存在.
如果f(x)在点x₀处可导,则lim_(hto0)(f(x_(0)+2h)-f(x_(0)))/(h)=()填空题(共15题,30.0分)题型说明:共15题,每
若极限 lim_(h to 0) (f(x_0 + 2h) - f(x_0))/(h) = (1)/(2),则导数值 f(x_0) = ( )。A. $-\fr
设函数 f(x) 在 x=0 处可导,则 lim_(h arrow 0) (f(2h)-f(-3h))/(h) = ( )A. $ -f'(0) $B. $ f
已知f(x)为可导函数且 (1)=-2, 则 lim _(harrow 0)dfrac (f(1-h)-f(1-2h))(2h)=
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[题目]设函数f (x)在 x=0 处连续,且 lim _(harrow 0)dfrac (f({h)^2)}({h)^2}=1,-|||-则 ()-|||-