Question

一些题求学霸解答，给下解题步骤谢谢

Answer 1

先给你前两题

追答

后面四道

Answer 2

2(1)，原式=1/0=∞
2(2),分子分母同时乘以√(x²+1)+x，原极限=lim x/(√(x²+1)+x)=lim 1/(√(1+1/x²)+1)=1/2
2(3)令t=2/x ->∞，则lim (1+x/2)^[(x-1)/x]=lim (1+1/t)^(1-t/2)
=lim (1+1/t)*[(1+1/t)^t}^(-1/2)=1*e^-1/2=e^-1/2
2(4) tanx-sinx=sinx(1/cosx-1)
=sinx(1-cosx)(1+cosx)/[cosx(1+cosx)]
=(sinx)^3/[cosx(1+cosx)]~x^3/2
所以原极限=lim x^3/2x²=0
2(5)分子分母同时除以n^4，易知极限为0
2(6)分子分母同时乘以√(3-x)+√(x+1),则原极限=
lim (3-x-x-1)/(x-1)(x+1)[√(3-x)+√(x+1)]
=lim -2/(x+1)[√(3-x)+√(x+1)]= -2/(2*2√2)= -√2/4

Answer 3

如下