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

[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

(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)}=-

2、设f(x)在区间[0,1]上可导, (1)=2(int )_(0)^dfrac (1{2)}(x)^2f(x)dx, 证明:存在 varepsilon in

(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)_

设f(x)为连续函数，则(int )_(0)^1f(dfrac (x)(2))dx等于( )．(int )_(0)^1f(dfrac (x)(2))dx设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9]设f(x)在[0,1]上连续, (0)=0, (int )_(0)^1f(x)dx=0.-|||-求证:存在 xi in (0,1), 使 (int )