Question

高数：求极限，一共五个，答案分别是2 cosα 1 e的负二分之三次方 二分之一，要过程，谢谢

Answer 1

lim(x->1)[√(5x-4) -√x]/(x-1)
=lim(x->1)(4x-4)/{ (x-1)[√(5x-4) +√x] }
=lim(x->1)4/[√(5x-4) +√x]
=4/(1+1)
=2

(2)
lim(x->α) (sinx- sinα)/(x- α) (0/0)
=lim(x->α) cosx
=cosα

(3)
lim(x-> ∞)[√(x^2+x) -√(x^2-x)]
=lim(x-> ∞)2x/[√(x^2+x) +√(x^2-x)]
=lim(x-> ∞)2/[√(1+1/x) +√(1-1/x)]
=2/(1+1)
=1

(4)
lim(x->∞) [( 3+x)/(6+x)]^ [(x-1)/2]
=lim(x->∞) [1 + 3/(6+x)]^ [(x-1)/2]

let
3/(6+x)= 1/y
3y= 6+x
x= 3y-6

lim(x->∞) [( 3+x)/(6+x)]^ [(x-1)/2]
=lim(x->∞) [1 + 3/(6+x)]^ [(x-1)/2]
=lim(y->∞) [1 + 1/y]^ [(3y-7)/2]
=e^(3/2)

追问

第二个是为什么？