Question

文科高数题目

lim[1/(x+1)-3/(x^3+1)] x趋向于-1 答案是-1 怎么做

lim[(e^2x-1)/x] x趋向于0 答案是2 怎么做
没学过罗比达法则..

Answer 1

(1)属于无穷大-无穷大型，先通分，变成0/0型，然后采用罗比达法则，得到结果-1.
(2)属于0/0型，最简单的办法是对分子采用等价无穷小代换：当x趋于0时，e^x等价于x；从而分子部分等价于2x,然后跟分母一约去x即得2.此题也可以直接运用罗比达法则，通过对分子分母分别求导数，结果也是2.

Answer 2

1/(x+1)-3/(x^3+1)
=[(x^2-x+1)-3]/(x^3+1)
=(x^2-x-2)/(x^3+1)
=(x-2)(x+1)/(x+1)(x^2-x+1)
=(x-2)/(x^2-x+1)
所以lim(x→-1)[1/(x+1)-3/(x^3+1)]
=lim(x→-1)(x-2)/(x^2-x+1)
=(-1-2)/(1+1+1)
=-1

当x趋向于0时，e^(2x)-1和x趋于0
是0/0型，可以用洛必达法则
lim(x→0)[(e^2x-1)/x]
=lim(x→0)2e^2x
=2

Answer 3

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

代入-1

所以为-3/3=-1

Answer 4

lim[1/(x+1)-3/(x^3+1)]
x^3+1=(x+1)(x^2-x+1)
原式=lim[(x^2-x+1-3)/(x^3+1)]=lim[(x-2)(x+1)/(x^3+1)
=lim[(x-2)/(x^2-x+1)]
当x趋向-1
原式=[(-1-3)/(1+1+1)]=-1

Answer 5

lim[1/(x+1)-3/(x^3+1)]=lim{1/(x+1)-3/[(x+1)(x^2-x+1)]}
=lim{(x^2-x+1)/[(x+1)(x^2-x+1)]-3/[(x+1)(x^2-x+1)]}
=lim{(x^2-x-2)/[(x+1)(x^2-x+1)]}
=lim{(x-2)(x+1)/[(x+1)(x^2-x+1)]}
=lim[(x-2)/(x^2-x+1)]
当x→-1,lim[(x-2)/(x^2-x+1)]→lim(-3/3)=-1

lim[(e^2x-1)/x] 不会做，工作时间长，很久没做高数题目了呵