Question

这几个极限怎么求，多谢了，写下详细步骤

Answer 1

(2)
lim(x->∞) (7x^4+5x-2)/(2x^2-x+1)
=lim(x->∞) (7x^2+5/x-2/x^2)/(2-1/x+1/x^2)
->∞
分子->∞
分母->2
(3)
lim(x->∞) (2x^2+5)/x^3
=lim(x->∞) (2+5/x^2)/x
=0
分子->2
分母->∞
(4)
lim(x->∞) [ (x^2+1)/(x+1) -(ax+b)] =0
lim(x->∞) [ (x^2+1) - ax^2-(a+b)x -b ]/(x+1) =0
lim(x->∞) [ (1-a)x^2 -(a+b)x +(1-b) ]/(x+1) =0
=>分子: x^2, x 的系数=0
1-a = 0 and a+b = 0
a=1 and b=-1
(5)
lim(x->2) (x^3+ax^2+b)/(x-2) =8 (0/0)
f(x)=x^3+ax^2+b
f(2) = 0
8+4a+b=0
4a+b= -8 (1)
lim(x->2) (x^3+ax^2+b)/(x-2) =8 (0/0)
lim(x->2) (3x^2+2ax) =8
12+4a=8
a=-1
from (1)
4a+b= -8
-4+b=-8
b=-4
(a,b)=(-1,-4)

追问

当分母需向0，分子分母都要除以最低次方么

追答

目的使得分子-》常数

追问

好的

Answer 2

图