(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. (D.)lim_(h to 0)(f(a)-f(a-h))/(h)存在.

(B)lim_(hto0)(1)/(h^2)f(sqrt(1+h^2)-1)存在.(1)/(h^2)f(tan h-sin h)存在. (D)lim_(hto

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

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

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

设 f(x) 在点 x_0 的邻域内存在，且 f(x_0) 为极大值，则 lim_(h to 0) (f(x_0 + 2h)- f(x_0))/(h) = (

(f(x)在 _{0)=(x)_(0) 处可导,且 lim _(harrow 0)dfrac (f({x)_(0)+2h)-f((x)_(0)-h)}(2h)=

6、单选-|||-f-|||-A .lim _(harrow 0)dfrac (f({x)_(0)+5h)-f((x)_(0)+2h)}(h)=f((x)_(0

若极限lim _(harrow 0)dfrac (f({x)_(0)+2h)-f((x)_(0))}(h)=dfrac (1)(2)，则导数值lim _(har

若函数f（x）在点x0处可导，则极限underset(lim)(h→0)(f（(x)_(0)+3h）-f（(x)_(0)-h）)/(2h)=（ ）A. 4f′（

[题目]已知 (3)=2, 则 lim _(harrow 0)dfrac (f(3-h)-f(3))(2h)= __