A. 4f′(x0)
B. 3f′(x0)
C. $\frac{1}{2}$f′(x0)
D. 2f′(x0)
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
(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)
(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)_
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
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
(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)}=-
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