6.下列各题中均假定f'(x0)存在,按照导数定义观察下列极限,指出A表-|||-示什么:-|||-(1) lim _(Delta xarrow 0)dfrac (f({x)_(0)-Delta x)-f((x)_(0))}(Delta x)=A ;-|||-(2) lim _(xarrow 0)dfrac (f(x))(x)=A ,其中 f(0)=0 ,且f'(0)存在;-|||-(3) lim _(harrow 0)dfrac (f({x)_(0)+h)-f((x)_(0)-h)}(h)=A .

1.已知f`(x0)存在,则由导数定义知, lim _(Delta xarrow 0)dfrac (f({x)_(0)-Delta x)-f((x)_(0))}

设f"(a)存在, (a)neq 0 ,则 lim _(xarrow a)[ dfrac (1)(f(a)(x-a))-dfrac (1)(f(x)-f(a))

19.若 ((x)_(0))=-2 ,则 lim _(Delta xarrow 0)dfrac (f({x)_(0)+Delta x)-f((x)_(0))}(

设y=f(x) 在x0处可导,且 ((x)_(0))=2, 则lim _(xarrow 0)dfrac (f({x)_(0)+2)x-f((x)_(0)-f(x

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)=(

C. lim _(xarrow 0)dfrac (f(x))(x) 不存在.-|||-f(0)=0 ^11(0)=2.-|||-D.f(0)是f(x)的极小值.

(B)若 lim _(xarrow 0)dfrac (f(x)+f(-x))(x) 存在,则 (0)=0.-|||-(C)若 lim _(xarrow 0)df

[题目]设f(x )具有二阶连续导数,且f(0)-|||-=0, lim _(xarrow 0)dfrac (f(x))(|x|)=1 则 ()-|||-A.f

6、单选-|||-f-|||-A .lim _(harrow 0)dfrac (f({x)_(0)+5h)-f((x)_(0)+2h)}(h)=f((x)_(0

设f(x)二阶可导, lim _(xarrow 0)dfrac (f(x))(x)=1 (1)=1, 证明:存在 xi in (0,1), 使得-|||-(xi