Question

第四题和过程谢谢！

Answer 1

lim(x->-8) [√(1-x) - 3]/[ 2+ x^(1/3) ] (0/0)
=lim(x->-8) { -1/[2√(1-x)] } /[ (1/掘迅3)x^(-2/3) ]
=(-3/2)lim(x->判神此-8) x^(2/3)/√(1-x)
=(-3/瞎枯2)(4/3)
=-2

追问

我没看懂第一步怎么变的

追答

lim(x->-8) [√(1-x) - 3]/[ 2+ x^(1/3) ] (0/0)
=lim(x->-8) d/dx [√(1-x) - 3]/ d/dx [ 2+ x^(1/3) ]
=lim(x->-8) { -1/[2√(1-x)] } /[ (1/3)x^(-2/3) ]

追问

如果不用洛必达法则解呢

追答

let
y= x+8
x->-8 , y->0

lim(x->-8) [√(1-x) - 3]/[ 2+ x^(1/3) ]
=lim(y->0) [√(9-y) - 3]/[ 2+ (y-8)^(1/3) ]

等价无穷小
√(9-y) ~ 3 -(1/6) y
(y-8)^(1/3) ~ -2 +(1/12)y

lim(x->-8) [√(1-x) - 3]/[ 2+ x^(1/3) ]
=lim(y->0) [√(9-y) - 3]/[ 2+ (y-8)^(1/3) ]
=lim(y->0) ( -1/6)y / [(1/12)y]
= -2

追问

谢谢