Question

limt(x+k/x-2k) 的x次方=8 求常数k的值 x→∞

Answer 1

(x+k)/(x-2k)

=1+3k/(x-2k)

1/a=3k/(x-2k)

x=3ak+2k

所以=(1+1/a)^(3ak+2k)

=[(1+1/a)^a]^(3k)*(1+1/a)^2k

极限=e^(3k)*1=8

k=(ln8)/3=ln2