[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 ()-|||-

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

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

(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)}=-

(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)= () ;-|||-(

例22 设f(x)在[0,1]上连续,证明:-|||-lim _(harrow {0)^+}(int )_(0)^1dfrac (h)({h)^2+(x)^2}

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

[题目]设f(x)在 x=0 处连续,且 lim _(xarrow 0)dfrac ((f(x)+1){x)^2}(x-sin x)=2,-|||-则曲线 =f

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)