设函数 f(x) 在 x=0 处可导，则 lim_(h arrow 0) (f(2h)-f(-3h))/(h) = （ ）

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′（

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

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

(1)若f(x)在 =(x)_(0) 处可导,则 () .-|||-(A) lim _(harrow 0)dfrac (f({x)_(0)+2h)-f((x)_

(6)设f(x)在x0处可导,则 lim _(harrow 0)dfrac (f({x)_(0)+h)-f((x)_(0)-h)}(h)= () ;-|||-(

(2)已知f(x)在 =(x)_(0) 处可导,且有 lim _(harrow 0)dfrac (2h)(f({x)_(0))-f((x)_(0)-4h)}=-

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

(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.

已知函数f(x)在点x0处可导,则下列极 限 中 () 等于导数值f`(x0).-|||-(A) lim _(harrow 0)dfrac (f({x)_(0)

[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 ()-|||-