[求助] 求一道高中的数学解答题解法已有1人参与

2楼: Originally posted by hylpy at 2016-07-28 10:03:00
解:由通项公式，可得:t_1=\frac{110a+90\times 15}{11a+15};
\therefore \lim_{a \to \infty  }t_1=\frac{110a+90\times 15}{11a+15}=10;
\Rightarrow \lim_{a \to \infty  }t_n=\frac{110a+90\times 15}{11a+15} ...