The value of `lim_(ntooo) (1)/(n)[e^(1/n)+e^(2/n)+...+e]` is

Answer

C

Explanation

`underset(ntooo)lim[(1)/(n)+(e^(1//n))/(n)+(e^(2//n))/(n)+...+(e^((n-1)//n))/(n)]`
`=underset(ntooo)lim[(1+e^(1//n)+(e^(1//n))^(2)+...+(e^(1//n))^(n-1))/(n)]`
`=underset(ntooo)lim(1.[(e^(1//n))-1])/(n(e^(1//n)-1))=(e-1)underset(ntooo)lim(1)/(((e^(1//n)-1)/(1//n)))`
`=(e-1)xx1=(e-1)`