求x趋向于无穷,求lim((2^1/x+3^1/x+…+100^1/x+1)/100)^100x的详细解答

Question

求x趋向于无穷,求lim((2^1/x+3^1/x+…+100^1/x+1)/100)^100x的详细解答

Answer 1

令t=1/x,
那么
原极限
=e^ lim(t->0) 100/t *ln[(2^t+……+100^t+1)/100]
=e^ lim(t->0) 100/t *ln[1+ (2^t+……+100^t+1 -100)/100]
那么显然
t趋于0时,
(2^t+……+100^t+1 -100)/100也趋于0,
所以
ln[1+ (2^t+……+100^t+1 -100)/100]
就等价于(2^t+……+100^t+1 -100)/100
于是
原极限
=e^ lim(t->0) 100/t * (2^t+……+100^t+1 -100)/100
=e^ lim(t->0) (2^t+……+100^t+1 -100)/t
而注意a^t-1等价于ln2,
所以
(2^t+……+100^t+1 -100)/t
=(1^t-1)/t +(2^t-1)/t +(3^t-1)/t +…+(100^t-1)/t
就等价于
ln1+ln2+ln3+…+ln100
那么就得到
原极限
=e^ lim(t->0) (2^t+……+100^t+1 -100)/t
=e^ (ln1+ln2+ln3+…+ln100)
=1 *2*3*…*100
=100!
所以
极限值为100!