lim[(sin x)^tan x],x趋向于Pai/2

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lim[(sinx)^tanx],x趋向于Pai/2lim[(sinx)^tanx],x趋向于Pai/2lim[(sinx)^tanx],x趋向于Pai/2lim(x->pi/2)[(sinx)^ta

lim[(sin x)^tan x],x趋向于Pai/2
lim[(sin x)^tan x],x趋向于Pai/2

lim[(sin x)^tan x],x趋向于Pai/2
lim(x->pi/2)[(sin x)^tan x]
=lim(x->pi/2){[1+(sinx-1)]^[1/(sinx-1)]}^[(sinx-1)/cotx]
=e^lim(x->pi/2)[(sinx-1)/cotx]
=e^lim(x->pi/2)[-cosx/(cscx)^2]
=e^(-0/1^2)
=1

应该是1
化成e的tan(x)ln(sin(x))次方
把tan(x)放到分母变成1/tan(X)
再用洛必答法则
得出{cos(x)^5}/{sin(x)^3}极限为0
所以为e的0次方
为1

1
用对数法

=lime^[tanxlnsinx]
=lime^[tanxln(1+sinx-1)]
=lime^[tanx(sinx-1)]
=lime^[sinx(sinx-1)/cosx]
=lime^[(sinx-1)/cosx]
=lime^[-cosx/sinx]
=e^0
=1

lim(sinx)^tanx＝e^lim[(lnsinx)/cotx] lim[(lnsinx)/cotx]=lim(-cosx*sinx)=0 洛必达法则求出来的，所以最后结果是e^0=1。希望能帮到你哈

Lim(sinx+cosx)^(sinx/cosx)=Lim[(sinx+cosx)^(1/cosx)]^sinx=e^sinx=1

lim x->0 (sin x-tan x)/sin (x^3) lim (x->0) [tan(tan x)-sin(sin x)]/(tan x -sin x) lim[(sin x)^tan x],x趋向于Pai/2 Lim*（sin x/tan x）x->0 lim (x^x-sin(x)^x)/(sin(x)sin(x)ln(1+tan(x))) x趋于零求 lim(tan x-sin x)/(sin x)^3 x趋于0的极限值详细步骤 lim(x→0)[tan(2x+x^3)/sin(x-x^2)] lim arc tan(x-sin(lnx)) n趋于无限大 lim(sin(aX)/tan(bX))(X趋近于0) lim(x(sin)^2) lim [(1+tan x)/(1+sin x)]^(1/x^3) x趋于零 lim(x→0)(e^tan x-e^sin x)/x^3,rt lim[x→1] tan(x-1)-sin(x-1) / x-1= lim(x→0) [1-cos(x^2)]/sin(x^2)tan(x^2)]=? lim(tan^3(3x)/(X^2sin(2x))(x趋近于0） lim(x->1/2)sin(1-2x)tanπx 当x趋近于pai lim[sin(3x)/tan(5x)] 当x趋于0时:lim (tan x - sin x )/x^3 lim [(a^x+b^x+c^x)/3]^(1/x) ( a,b,c > 0 )