设f`(x)连续, (0)=0, '(0)=2, 极限 lim _(xarrow 0)dfrac ({int )_(0)^xln cos (x-t)dt}(sqrt {1+{f)^2(x)}-1} 为()

[例1.20]设f`(x)连续, (0)=0, (0)neq 0, 求 lim _(xarrow 0)dfrac ({int )_(0)^(x^2)f((x)^

求下列极限: lim _(xarrow 0)[ dfrac ({int )_(0)^xsqrt (1+{t)^2}dt}(x)+dfrac ({int )_(0

求极限lim _(xarrow 0)dfrac ({int )_(0)^2xln (1+t)dt}(xsin x)求极限

设f(x)连续, varphi (x)=(int )_(0)^1f(xt)dt, 且 lim _(xarrow 0)dfrac (f(x))(x)=A设f(x)

求极限lim _(xarrow 0)dfrac ({int )_(0)^x(1-cos t)dt}({x)^3}= (ost)dt/=()。求极限。

41.求极限 lim _(xarrow 0)dfrac ({int )_(0)^xln (t+(e)^t)dt}(1-cos x)

lim _(xarrow 0)dfrac (x{int )_(0)^x((e)^2t-1)dt}(ln (1+{x)^3)}。。

(2)设函数f(x)在区间 (-1,1) 内有定义,且 lim _(xarrow 0)f(x)=0, 则-|||-(A)当 lim _(xarrow 0)dfr

.求极限 lim _(xarrow 0)dfrac (x-tan x)(xln (1+{x)^2)}

lim _(xarrow 0)dfrac ({int )_(x)^0((e)^t+(e)^-t-2)dt}(1-cos x) ()-|||-__=（）=（）A.