limx+0xcotx解题步骤

刁贝晏3801xcotx 当x趋近于0是的极限时多少, -
权浦怖14720237485 ______[答案] (x->0) lim xcotx =(x->0) lim xcosx/sinx =(x->0) lim cosx/ lim (sinx/x) =1/1 =1

刁贝晏3801lim(XcotX)在X趋近于0的极限值. -
权浦怖14720237485 ______ lim(x→0) XcotX =lim(x→0) X/tanX =1

刁贝晏3801高中题目,求x*cotx当x→0时的极限 -
权浦怖14720237485 ______ xcotx=xcosx/sinx 一个重要的极限不知道你们学没学过lim_{x→0}(sinx/x)=1 由此,所求极限为1

刁贝晏3801为什么lim(x→0)xcotx=1,cot0不是不存在的吗 -
权浦怖14720237485 ______ cotx = 1/tanx 所以,上式中的极限可以变换为: =lim x/tanx 这是一个 0/0 型的极限,可以使用罗必塔法则: =lim (x)'/(tanx)' =lim 1/sec²x =lim 1/sec²0 =lim 1/1² =1

刁贝晏3801当x趋近于0时,xcotx的极限 -
权浦怖14720237485 ______ =lim(x/tanx) =lim(x/x) =1

刁贝晏3801问:lim其中x—>0+时求limx^(cotx)=?要有步骤 -
权浦怖14720237485 ______ =e^lim cotx·lnx =e^lim (lnx/tanx) =e^lim (lnx/x) =e^lim (1/x) →+∞

刁贝晏3801x趋向0 lim xcotx -
权浦怖14720237485 ______[答案] 1 xcotX=xcosx/sinx=cosx*(x/sinx) x→0时,cosX=1 下面求lim x→0 sinx/x的极限 用夹挤定理 由sinx

刁贝晏3801lim x趋近0 cot(x)求xcotx -
权浦怖14720237485 ______[答案] lim x趋近0 xcot(x) =lim x趋近0 xcos(x)/sinx =lim x趋近0 cos(x) =1

刁贝晏3801limx→0cotx(1sinx?1x)= - ----- -
权浦怖14720237485 ______ 因为:cotx(1 sinx ?1 x )= cosx sinx x?sinx xsinx ,且当x→0时,x-sinx~1 6 x3,cosx→1,所以:lim x→0 cotx(1 sinx ?1 x )= lim x→0 cosx sinx x?sinx xsinx = lim x→0 x?sinx x3 = lim x→0 1 6 x3 x3 =1 6 . 故答案为:1 6 .

刁贝晏3801lim(x→0)SinωX/X的极限是ω吧,需要个步骤,谢谢! -
权浦怖14720237485 ______ 答: 1. lim(x→0)SinωX/X =lim(x→0)SinωX/ωX*ω =ω 2. lim(x→0)tan3X/X =lim(x→0)sin3X/(Xcos3X) =lim(x→0)sin3X/X =lim(x→0)sin3X/3X*3 =3 3. lim(x→0)xcotx =lim(x→0)x/tanx 同上 =1