Question

求大神帮忙解答一下这个题

Answer 1

lim(x->∞) [(x-1)/(x+1)]^x
=lim(x->∞) [1 - 2/(x+1)]^x
=e^(-2)
lim(x->∞) [ cos(c/x) ]^(x^2)
=lim(x->∞) [ 1- (1/2)(c/x)^2 ]^(x^2)
=lim(x->∞) [ 1- (1/2)c^2(1/x^2) ]^(x^2)
=e^(-(1/2)c^2)
-2 = -(1/2)c^2
c^2=4
c=2

Answer 2

A = lim<x→∞>[(x-1)/(x+1)]^x = lim<x→∞>{[1-2/(x+1)]^[-(x+1)/2]}^[-2x/(x+1)]
= lim<x→∞>e^[-2x/(x+1)] = e^lim<x→∞>[-2/(1+1/x)] = e^(-2)
记 B = [cos(c/x)]^x^2, lnB = x^2lncos(c/x) = lncos(c/x)/(1/x^2)
lim<x→∞>lnB = lim<x→∞>lncos(c/x)/(1/x^2) (0/0)
= lim<x→∞>sec(c/x)sin(c/x)(c/x^2)/(-2/x^3) = lim<x→∞>csin(c/x)/(-2/x)
= lim<x→∞>c(c/x)/(-2/x) = -c^2/2, B = e^(-c^2/2)
A = B, 2 = c^2/2, c = 2, c = -2(舍去)