(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
(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)}=-
(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)
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
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′(
设 函数 f ( x ) 在 x = 0 处可导,且lim _(xarrow 0)dfrac (f(2x)-f(0))(ln (1+3x))=1,则f(0)=(
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