Question

大一高等数学求极限

1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限
2 cotx -1/x当x趋于0的极限
3 （x^3+x^2+x+1)^1/3-x 当x趋于∞时的极限

Answer 1

1.[㏑(x-π/2)]/tan x 当x趋于π/2时的极限

=lim(x->π/2)1/(x-π/2)/sec²x
=lim(x->π/2)cos²x/(x-π/2)
=lim(x->π/2)2cosx(-sinx)/1
=0
2.
lim(x->0)cotx-1/x

=lim(x->0)xosx/sinx-1/x
=lim(x->0)(xcosx-sinx)/xsinx
=lim(x->0)(xcosx-sinx)/x²
=lim(x->0)(cosx-xsinx-cosx)/2x
=lim(x->0)(-xsinx)/2x
=lim(x->0)(-sinx)/2
=0
3.（x^3+x^2+x+1)^1/3-x
=lim(x->∞)e^ln（x^3+x^2+x+1)^1/3-x
=lim(x->∞)e^[ln（x^3+x^2+x+1)]/(3-x)
=e^lim(x->∞)1/（x^3+x^2+x+1) *(3x²+2x+1)/(-1)
=e^lim(x->∞)-(3x²+2x+1)/（x^3+x^2+x+1)
=e^lim(x->∞)-(6x+2)/（3x^2+2x+1)
=e^lim(x->∞)-(6)/（6x+2)
=e^0
=1

追问

=lim(x->∞)e^[ln（x^3+x^2+x+1)]/(3-x)

这一步有问题吧？后面的-x是脱离e^……的呀，怎么变的？？

追答

ln（x^3+x^2+x+1)^1/3-x

=1/(3-x)ln（x^3+x^2+x+1)
=[ln（x^3+x^2+x+1)]/(3-x)

追问

抱歉啊，我写的好像有点问题
（x^3+x^2+x+1)^(1/3)-x

追答

这个题目，我给你思路，自己做吧
令x=1/t
x->∞，t->0