证明等式 arctan x=arcsin dfrac (x)(sqrt {1+{x)^2}} , in (-infty ,+infty

lim _(xarrow infty )(x)^2(sqrt (dfrac {1+{x)^2}({x)^2}}-1)= __

2.求极限 lim _(xarrow +infty )arcsin dfrac (x)(sqrt {{x)^2+1}+x}

求极限 lim _(xarrow +infty )((x+sqrt {1+{x)^2})}^dfrac (1{x)}

设=dfrac (arcsin x)(sqrt {1-{x)^2}}（1）证明：=dfrac (arcsin x)(sqrt {1-{x)^2}}（2）求=df

设(x)=dfrac (1)(1+{e)^x}+1,xin (-infty ,+infty )且 (x)=dfrac (1)(1+{e)^x}+1,xin (-

(int )_(-infty )^+infty (x)^2(e)^-a(x^2)dx=dfrac (1)(2)sqrt (dfrac {pi )({a)^3}}

(2) lim _(xarrow infty )dfrac (arctan x)(x).

、证明:当 -1lt xlt 0 时, arcsin sqrt (1-{x)^2}-arctan dfrac (x)(sqrt {1-{x)^2}}=dfrac

()-|||-(2) (int )_(-infty )^+infty dfrac (2x)(1+{x)^2}dx=lim _(aarrow +infty )(i

求极限lim _(xarrow +infty )x(sqrt (1+{x)^2}-x)。求极限。