已知 '(0)=3, 则 lim _(harrow 0)dfrac (f(-h)-f(0))(3h)=

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

设函数f(x)满足lim _(harrow 0)dfrac (1)(h)[ f(5-dfrac (1)(3)h)-f(5)] =2,则lim _(harrow

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

16.若 (x)=ln (1+(x)^2), 则 lim _(harrow 0)dfrac (f(3)-f(3-h))(h)= __

(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)}=-

已知f(x)为可导函数且 (1)=-2, 则 lim _(harrow 0)dfrac (f(1-h)-f(1-2h))(2h)=

注:已知 (0)=0, 则下列说法中与函数f(x)在点 x=0 处可导等价的是 ()-|||-(A)极限 lim _(harrow 0)dfrac (f(({e

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