由条件知：tbn+1=2b(n+1),且t≠2.可得
b(n+1)+1/(t-2)=(t/2)[bn+1/(t-2)].
由f(b)≠g(b),t≠2,t≠0,可知b+1/(t-2)≠0,t/2≠0,
所以{bn+1/(t-2)}是首项为b+1/(t-2),公比为t/2的等比数列.
bn+1/(t-2)=[b+1/(t-2)](t/2)^(n-1),即bn=[b+1/(t-2)](t/2)^(n-1)-1/(t-2)
由an=2b(n+1)可知,若liman(n→∞）存在,则limbn(n→∞）存在.
于是可得0＜|t/2|＜1,故-2＜t＜0或0＜t＜2
liman(n→∞）=2limbn(n→∞）=2/(t-2).