Question

高数！！！

Answer 1

nan=∫<上n/n+1，下0>d[1+x^n]^(3/2)
={1+[n/(n+1)]^n}^(3/2)-1
lim(n->+无穷)nan
=lim(n->+无穷){1+[n/(n+1)]^n}^(3/2)-1
={1+lim(n->+无穷)[n/(n+1)]^n}^(3/2)-1
现求lim(n->+无穷)[n/(n+1)]^n即可
设y=[n/(n+1)]^n
lny=nln[n/(n+1)]
lim(n->+无穷)lny=lim(n->+无穷)nln[n/(n+1)]
=lim(n->+无穷)ln[n/(n+1)]/(1/n)
由罗必达 =lim(n->+无穷)[-n/(n+1)]
=-1
=>lim(n->+无穷)y=e^(-1)
=>lim(n->+无穷){1+[n/(n+1)]^n}^(3/2)-1
=[1+e^(-1)]^(3/2)-1
最终答案为（B）
希望能帮到你!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

nan=3/2*积分（从0到n/(n+1)）根号(1+x^n)d(1+x^n)
=(1+x^n)^(3/2)|上限n/(n+1)下限0
=（1+(n/n+1)^n）^(3/2)--1
趋于(1+1/e)^(3/2)--1，选B