若数列{an}满足a1=根号6 a(n+1)=根号下an+6 (n∈N*) 如果lim an 存在,求lim an的值[数列极限]

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liman=liman+1 an+1=根滑孝庆号下慎前an+6
即liman+1=根信握号下liman+1 +6
liman+1=3或-2 -2舍去（显然an>0)
所以lim=3

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