tanx除以tan3x极限洛必达

郜呼淑1968为什么tanx/tan3x不能用等价无穷小来做?如图用的是洛必达法则. -
胡潘行13465435034 ______ 因为当x趋近二分之派时,tanx是无穷大的,不是无穷小

郜呼淑1968lim(x趋于π/2)(tanx/tan3x)求极限 -
胡潘行13465435034 ______[答案] lim(x→π/2)tan3x/tanx =lim(x→π/2) 3sec^2(3x)/sec^2x =lim(x→π/2) 3cos^2x/cosx^2(3x) =lim(y→0) 3sin^2y/sin^2(3y) (令y=π/2-x) =lim(y→0) 3y^2/(3y)^2 =3/9 =1/3

郜呼淑1968用洛必达法则当x - >派/2时,tanx/tan3x等于多少?我怎么算都得1啊!一楼你的明显是错的,用洛必达法则 -
胡潘行13465435034 ______[答案] lim(x->π/2)(tanx/tan3x)=lim(x->π/2)(ctan3x/ctanx) =lim(x->π/2)(ctan'3x/ctan'x) =lim(x->π/2)((-1/sin^2(3x)*3/(-1/sin^2(x)) =3sin^2(π/2)/sin^2(3π/2) =3

郜呼淑1968x→pi/2 lim tanx/tan3x=? -
胡潘行13465435034 ______[答案] 设u=x-π/2 lim(x→π/2)tanx/tan3x =lim(u→0)tan(u+π/2)/tan(3u+3π/2) =limcot(-u)/cot(-3u) =limtan(-3u)/tan(-u) =3

郜呼淑1968高数之极限 -
胡潘行13465435034 ______ 1.原式=(sinx/cosx)/(sin3x/cos3x)=(cos3x)/(cosx)=(4cos^3x-3cosx)/cosx=4cos^2x-1 x→π/2 原式=-12.设t=1/x^2 原式=e^t/t, t→正无穷 原式上下罗比达法则得:e^t趋于正无穷

郜呼淑1968极限lim( x→π/2) xtanx / tan3x -
胡潘行13465435034 ______[答案] 令t=x-π/2,原极限= lim(t-->0) (t+π/2)tan(3t)/tant= lim(t-->0) (t+π/2)(3t)/t= 3π/2.

郜呼淑1968解方程tanx=tan3x
胡潘行13465435034 ______ x=k¤/2 (¤代表圆周率3.1415926…因为手机打不出那个符号,所以…)

郜呼淑1968化简 1+tanx除以1 - tanx -
胡潘行13465435034 ______[答案] 1+tanx除以1-tanx =tan45°+tanx除以1-tan45°tanx =tan(45°+x)

郜呼淑1968化简tanxtan2x/tan2x - tanx,所得结果是 -
胡潘行13465435034 ______[答案] tanxtan2x/tan2x-tanx=2tan^2x/1-tan^2x/2tanx-tanx(1-tan^2x)/1-tan^2=2tan^2x/2tanx-tanx(1-tan^2x)=2tan^2x/tanx+tan^3x=2tanx/1+tan^2x=(2sinx/cosx)/(1/cos^2x)=2sinx/cosx*cos^2x=2sinxcosx=sin2x

郜呼淑1968已知f(tanx)=tan3x分之一减cos3x,求f( - 1) -
胡潘行13465435034 ______[答案] tan3x=tan(x+2x)=(tanx+tan2x)/(1-tanxtan2x)=[tanx+2tanx/(1-(tanx)^2)]/[1-2tanx^2/(1-(tanx)^2)]=[tanx-(tanx)^3+2tanx]/[1-3(tanx)^2]cos3x=cos(x+2x)=cosxsin2x+sinxcos2x=2sinx(cosx)^2+sinx(1-2(sinx)^2=2sin...