7、判断题 lim _(x arrow 0) ( int _(0)^x sin tdt)/(x)=1 。()

极限lim _(x arrow 0) x sin (1)/(x^2)=（ ）A. 1；B. $\infty$；C. 不存在.D. 0；

求lim _(arrow 0)dfrac (x-sin x)({(arctan x))^3}=lim _(arrow 0)dfrac (x-sin x)({(a

极限lim _(xarrow 0)dfrac ({int )_(0)^(x^2)(e)^tdt}(2x)=()A、 lim _(xarrow 0)dfrac (

计算极限：lim _(xarrow 0)[ dfrac ({{int )_(0)}^x(e)^-tcos tdt}({ln )^2(1+x)}-dfrac (1

(1)已知函数 (x)=(int )_(0)^x(e)^cos xdt g(x)=(int )_(0)^sin x(e)^tdt ,则 ()-|||-( )

求极限lim _(xarrow 0)dfrac ({int )_(0)^x((e)^t-1)cos tdt}(xarctan x)-|||-__求极限

极限 lim_(x arrow 0) ( (2 + e^frac(1)/(x))(1 + e^(2)/(x)) + (sin x)/(|ln(1+x)|) )

1.[判断题] 判断：设sin y+int_(0)^x^(2)cos tdt=e^y，则(dy)/(dx)=(2xcos x^2)/(cos y-e^y).（）

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

lim _(x arrow 0)(x+2 e^x)^(1)/(x-1) = ( ).A. 1B. $\frac{1}{2}$C. $\frac{1}{4}$D.