求x→0时 lim(arctanx)/x设x=tan(t) x→0时t→0 故为lim(t/tant)=1问 为什么 x→0时t→0 x→0时可以是t趋紧于派或二派啊?

求lim(x→0)arctanx/x的极限, 求极限lim(x→0)sinxsin(1/x);lim(x→∞)(arctanx/x) lim(x→0 )(arctanx/x)^(1/x^2)求极限求lim(x→0 )(arctanx/x)^(1/x^2) 求lim(x→0) (1/x^2)ln(arctanx/x) 求这个极限时,能不能直接用arctanx~x代入?把 ln(arctanx/x)变成ln(x/x)?理由. lim(x→+∞)(2/π*arctanx)^x求极限 求极限lim(2/π*arctanx)^x x→∞ 求极限解题lim(x→∞)=arctanx/x 求lim x趋向于0(arctanx)/(x^+1) 求lim x趋向于0(arctanx)/(x^2+1) lim(x+arctanx)/(x-arctanx) (x→∞) lim x→-∞ arctanx/2求极限 一.x---->0时,证明lim(arctanx)/x=1 求x趋近于0时的极限：lim(1/(arctanx)^2-1/x^2) lim(n趋近于0)(arctanx)/x lim(arctanx/sinx) x－>0 lim(x→0) (1-cos(1-cosx))/(arctanx^2)^2求极限 lim当x→0,(e∧x+ln(1-x)-1)/x-arctanx 用洛必达法则求极限 Lim(arctanx)ln(1-x)求极限是为什么x默认趋近于0+×→0即Limln(1-x)ln(arctanx)×→0+