lim(x→0) (1-cos(1-cosx))/(arctanx^2)^2求极限
用洛必达法则求极限 1,lim(x→0)arctanx-x/sinx^3 2,lim(x→0)lncosax/lncos
求1、lim(x→∞)10^1/x的极限 2、lim(x→1)arctanx的极限,
求极限lim(x→∞)(1+1/x)(2-((x+1)/x^2)*arctanx)
求极限lim(x→0)sinxsin(1/x);lim(x→∞)(arctanx/x)
limx→1(1-x)^(cosxπ/2)求极限lim(2/π arctanx)^x 其中x趋向于正无穷大
求极限 x 趋于0 lim(cosx)^1/(x^2)
求极限lim(x→0)(1-根号cosx)/[x(1-cos根号x)]
用无穷小量的性质求下列极限,1,x趋向于0,limx^2cos(1/x) 2,x趋向于无穷大,lim(arctanx/x
用洛必达法则求下列函数的极限.lim((兀/2-arctanx)/(1/x)) x→∞
求x趋近于0时的极限:lim(1/(arctanx)^2-1/x^2)
lim(x→+∞)(2/π*arctanx)^x求极限
求极限lim(1-√cosx)/(1-cos√x) (x→0+)