Question

怎么求极限，怎么想都想不出来怎么用两个重要极限

怎么求极限，怎么想都想不出来怎么用两个重要极限1和3小题

Answer 1

(1)sin(x²-1)~x²-1,除以x-1之后就得到x+1=2了呀
(3)底数化为1-2/(x+1),指数化为-(x+1)/2*(-4)-3
于是[1-2/(x+1)]^[-(x+1)/2*(-4)-3]
={[1-2/(x+1)]^[-(x+1)/2}^(-4)÷[1-2/(x+1)]^(-3)
=e^(-4)÷1^(-3)
=e^(-4)

Answer 2

(1)

lim(x->1) sin(x^2-1)/(x-1) (0/0)

=lim(x->1) 2x.cos(x^2-1)

=2

(3)

(x-1)/(x+1) = 1 - 2/(x+1)

let

1/y = 2/(x+1)

2y = x+1

x->∞ , y->∞

lim(x->∞) [(x-1)/(x+1)]^(2x-1)

=lim(x->∞) [1 - 2/(x+1) ]^(2x-1)

=lim(y->∞) [1 - 1/y ]^(4y-2)

=lim(y->∞) [1 - 1/y ]^(4y)

=e^(-4)