∫(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呵呵，還有個問題，lim(x→0,y→0)[(xy+5)^1/2-5]/xy=?lim(x→0,y→0)[(xy+5)^1/2-5]/xy設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)可是好像老師給的答案是負無窮誒。。lim(x→0,y→0)[(xy+5)^1/2-5]/xy設u=xy ,x→0,y→0,u→0=lim(u→0)[(u+5)^1/2-5]/u(u+5)^(1/2)-5<0，不能用羅比達法則1/u趨向無窮=負無窮