(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)_

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

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

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)

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

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

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

[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 ()-|||-