11.已知 f(0)=0 f(0)=2 ,则 lim _(narrow infty )([ f(dfrac {1)({n)^2})-dfrac (1)({n)^
dfrac (1)(3)f(x) C. -3f(x) D. -dfrac (1)(3)f(x)
设f"(a)存在, (a)neq 0 ,则 lim _(xarrow a)[ dfrac (1)(f(a)(x-a))-dfrac (1)(f(x)-f(a))
6、单选-|||-f-|||-A .lim _(harrow 0)dfrac (f({x)_(0)+5h)-f((x)_(0)+2h)}(h)=f((x)_(0
f(0)=0-|||-D. f(0)=1
14.设f(x)在闭区间[0,2]上二阶可导,且f(0)=0,f(1)=1,f(2)=-1,证明:至少存在一点ξ∈(0,2),使得f(ξ)+2ξf(ξ)+ξf(
。-|||-8.设 f(x)= ,xgt 0, 0,x=0, dfrac {1-cos {x)^2}(x),xlt 0, .-|||-x=0,求f(x),
设函数f(x)具有2阶导数,且f(0)=f(1),|f(x)|leq1。证明:(1) 当xin(0,1)时,|f(x)-f(0)(1-x)-f(1)x|leq(
设 f(1)=1 ,则 =lim _(xarrow 0)dfrac (f(1+x)-f(1-2sin x))(x+2sin x)= .__
10、甲四-|||-设 y=f(x) 由参数方程 =-|||-(453)-|||-A .dfrac {f({e)^3t-1)}(f(2t))-|||-B .