解∫sinx^2cosx^5dx,y=(sin1/x+1/x),x→0时的极限,lim(1+sinx)^1/x,x→0,

来源：学生作业帮编辑：大师作文网作业帮分类：数学作业时间：2024/11/12 11:30:30

解∫sinx^2cosx^5dx,y=(sin1/x+1/x),x→0时的极限,lim(1+sinx)^1/x,x→0,lim(x^2-4)/(x^2+2x-8),x→2;
求y=ln[(1-sinx)/(1+sinx)]的导数.最后一题要求详解~~

∫(sinx)^2(cosx)^5dx
=∫(sinx)^2(1-(sinx)^2)cosxdx
=∫(sinx)^2dsinx-∫(sinx)^4dsinx
=(sinx)^3/3 -(sinx)^5/5 +C
lim(x→0)(sin1/x+1/x)
x→0 1/x→∞
lim(x→0)sin(1/x)不存在
lim(x→0) (1+sinx)^(1/x) =lim( x→0)[(1+sinx)^(1/sinx)]^(sinx/x)
lim(x→0)sinx/x=1
=lim(sinx→0,x→0)(1+sinx)^(1/sinx)^(sinx/x)
=e
lim(x→2)(x^2-4)/(x^2+2x-8)
=lim(x→2)(x^2-4)'/(x^2+2x-8)'
=lim(x→2)(2x)/(2x+2)
=2/3
y=ln(1-sinx)/(1+sinx)
(1-sinx)/(1+sinx)=2/(1+sinx)-1 [(1-sinx)/(1+sinx)]'= -2(sinx)'/(1+sinx)^2= -2cosx/(1+sinx)^2
y'=[(1-sinx)/(1+sinx)]' / [(1-sinx)/(1+sinx)]
=[ - cosx/(1+sinx)^2 ] *[(1+sinx)/(1-sinx)]
= -2cosx/[(1+sinx)(1-sinx)]
=-2cosx/(cosx)^2= -2/cosx
再问：呵呵，还有个问题，lim(x→0,y→0)[(xy+5)^1/2-5]/xy=?
再答： lim(x→0,y→0)[(xy+5)^1/2-5]/xy 设u=xy ,x→0,y→0,u→0 =lim(u→0)[(u+5)^1/2-5]/u =lim(u→0)[(u+5)^1/2-5]'/u' =lim(u→0)(1/2)*[1/(u+5)^1/2)] =1/(2√5)
再问：可是好像老师给的答案是负无穷诶。。
再答： lim(x→0,y→0)[(xy+5)^1/2-5]/xy 设u=xy ,x→0,y→0,u→0 =lim(u→0)[(u+5)^1/2-5]/u (u+5)^(1/2)-5

解∫sinx^2cosx^5dx,y=(sin1/x+1/x),x→0时的极限,lim(1+sinx)^1/x,x→0, lim x趋于0 (sinx+x^2sin1/x)/[(1+cosx)ln(1+x)] 求lim(x→0)[(x^2sin1/x)/sinx] 求极限 lim x-0 1-cosx / x-sinx 求Lim(x→0)(sinx/x)^(cosx/1-cosx) Lim,x-0,(1/sinx)*(1/x-cosx/sinx)=? lim ln(1+x)/x^2(x趋于0)和lim x^2sin1/x/sinx(x趋于0)的极限是多少? lim[x→0] (1-sinx)^(1/x)的极限证明：极限lim(x→0)(sinx/x)=1 . lim(x→0) x-sinx/x+sinx的极限 x→0,lim(1-cosx)[x-ln(1+tanx)]/sinx^4的极限求极限lim(x→0)][ln(1+2x^2)+n sinx]/(1-cosx)