(1) lim _(xarrow 0)dfrac ({({int )_(0)^x(e)^(t^2)dt)}^2}({int )_(0)^xt(e)^2(t^2)dt} :

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

lim _(xarrow 0)(int )_(0)^xdfrac (1)({x)^3}((e)^-(t^2)-1)dt= __

设(x)=(int )_(0)^xt(e)^-(t^2)dt，则 A、(x)=(int )_(0)^xt(e)^-(t^2)dt的是单调下降的函数 B、(x)=

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

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

计算极限lim _(xarrow 0)dfrac ({x)^2sin 2x}(x-{int )_(0)^x}(e)^(t^2)dt.计算极限.

求下列极限：(1)lim _(xarrow 0)((int )_(0)^xcos (t)^2dt)/(x)；(2)underset(xto 0)(mathrm{

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

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

1.求极限 lim _(xarrow 0)dfrac ({int )_(0)^x(tan t-sin t)dt}(({e)^(x^2)-1)cdot ln (1