Question

Question #90ab5

Answer 1

`lim_(n->oo)a_n=2`

Explanation:

`a_(n+1)=sqrt(2+a_n)`

Assuming that `a_n` exists when `n->oo`, then
`lim_(n->oo)a_(n+1)=lim_(n->oo)sqrt(2+a_n)`
`lim_(n->oo)a_n=lim_(n->oo)sqrt(2+a_n)`
`lim_(n->oo)a_n^2=lim_(n->oo)2+a_n`
`lim_(n->oo)a_n^2-a_n-2=0`
`lim_(n->oo)a_n=(1±sqrt(1^2+4*2))/2=(1±3)/2`

However, `a_n>0` for all `n`. Thus,
`lim_(n->oo)a_n=2`