Question

求x趋向于无穷,求lim((2^1/x+3^1/x+…+100^1/x+1)/100)^100x的

Answer 1

问：求x趋向于无穷,求lim((2^1/x+3^1/x+…+100^1/x+1)/100)^100x的

答：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!