# Limit of sequence an an=radical n order(3^n+2010^n) n tend to `oo`

*print*Print*list*Cite

### 1 Answer

A general term of the sequence is `a_n=(3^n+2010^n)^(1/n)`

we want `lim_(n->oo)a_n=?`

we can write

`a_n={2010^n(1+(3/2010)^n}^(1/n)`

`=2010(1+(1/670)^n)^(1/n)`

`lim_{n->oo}a_n=2010`

because

`lim_{n->oo}(1+1/670^n)=1`

Thus answer is `lim_(n->oo)a_n=2010`

**Sources:**