夹逼定理求极限问题…求解n次根号下sin^2 1+sin^2 2+……+sin^2 n 求极限,用夹逼定理,刚学高数,

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夹逼定理求极限问题…求解
n次根号下sin^2 1+sin^2 2+……+sin^2 n 求极限,用夹逼定理,刚学高数,

∵ sin²1+sin²2+……+sin²n > sin²1
sin²1+sin²2+……+sin²n < 1+1+.+1 = n
∴ (sin²1)^(1/n) < (sin²1+sin²2+……+sin²n)^(1/n) < n^(1/n)
设 un = (sin²1+sin²2+……+sin²n)^(1/n),对un取对数,则有:
2ln(sin1)/n < ln(un) < ln(n)/n
(n→∞)lim[2ln(sin1)/n] < (n→∞)lim[ln(un)] < (n→∞)lim[ln(n)/n]
∵ (n→∞)lim[2ln(sin1)/n] = 0; (n→∞)lim[ln(n)/n] =0
∴ 根据夹逼定理有:
(n→∞)lim[ln(un)] = 0 ==> (n→∞) lim(un) = e^0 =1;
结论:
(n→∞) lim[(sin²1+sin²2+……+sin²n)^(1/n)] =1