求极限：sin x^tan x(x→π/2)

lim(sinx)^tanx (x→π/2) 求极限

x→π/2lim (sinx)^tanx换元:t=π/2-x,x=π/2-t=lim(t→0) [sin(π/2-t)]^tan(π/2-t)=lim (cost)^cott=lim e^ln (cost)^cott根据复合函数的极限运算:lim(x→x0) f(g(x))=f(lim(x→x0) g(x)) =e^ .

lim(sinx)^tanx (x→π/2) 求极限值

解法一:∵lim(x->π/2)[(sinx-1)tanx]=lim(x->π/2){[(sinx-1)/cosx]sinx} =lim(x->π/2)[(sinx-1)/cosx]*lim(x->π/2)(sinx) =lim(x->π/2){[sin(x/2)-cos(x/2)]/[cos(x/2)+sin(x/2)]}*1 =0*1 =0 lim.

lim(sinx)^tanx,x→π/2要怎么求极限啊?求大神指点,不要百度!!!看

lim(x→π)(sinx/tanx)=lim(x→π)sin(π-x)/tan(x-π)=lim(x→π)-sin(π-x)/tan(π-x)=lim(t→0)-sint/tant(令t=π-x)=lim(t→0)-t/t=-1

求极限:sin x^tan x(x→π)

这个题目有问题,x→π+时,sinx是负值,无法取对数的.

求极限(sinx)^tanx,当x趋向与π/2的时候

我只想说,你给出的解题过程是错的,而且是没有道理的x→π/2lim (sinx)^tanx换元:t=π/2-x,x=π/2-t=lim(t→0) [sin(π/2-t)]^tan(π/2-t)=lim (cost)^cott=lim e^ln (cost)^cott根据复合.

求lim(x趋近于π)(π - x)tan(x/2)的极限?

π-x 趋近于0; tan(x/2)趋近于∞;lim(x趋近于π)(π-x)tan(x/2) =lim(x趋近于π)(π-x)/ (cos(x/2)/sin(x/2)) =(分子分母同时求导)lim(x趋近于π)2*(-1)/(-1/(sin x/2)^2 )=lim(x趋近于π)(-2)*(sin π2)^2 / (-1) =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

求极限:lim(1 - x)tanπx/2(x→1)

即(1-x)sin(πx/2)/cos(πx/2) x趋于1,那么 此时sin(πx/2)趋于1,1-x趋于0 而cos(πx/2)=sin(π/2-πx/2)=sinπ/2(1-x) 由重要极限得到(1-x)/ sinπ/2(1-x)=2/π *[π/2(1-x)]/sinπ/2(1-x) 后者趋于1,于是极限值为2/π

求极限lim(x→1)[(1 - x)tan(πx/2)]

tan(πx/2)=cot(π/2-πx/2)=1/tan(π/2-πx/2)等价于1/(π/2-πx/2) ∴原式=lim(x→1)(1-x)/(π/2-πx/2)=2/π

x→∏/2时, sin x 的 tan x 次方的极限是多少

设y=sinx^(tanx),则lny=tanx*ln(sinx)=sinx/cosx*ln[1+(sinx-1)]x→π/2时,sinx→1,ln[1+(sinx-1)]等价于sinx-1,所以lim(x→π/2) lny=lim(x→π/2) tanx*ln(sinx)=lim(x→π/2) 1/cosx*(sinx-1) 洛必达法则=lim(x→π/2) cosx/(-sinx)=0所以,lim(x→π/2) sinx^(tanx)=e^(0)=1