`a_n = (5n^2)/(n^2+2)`

To determine the limit of this sequence, let n approach infinity.

`lim_(n->oo) a_n`

`=lim _(n->oo) (5n^2)/(n^2+2)`

To solve, factor out the `n^2` in the denominator.

`=lim_(n->oo) (5n^2)/(n^2(1+2/n^2))`

Cancel the common factor.

`= lim_(n->oo) 5/(1+2/n^2)`

Then, apply the rule `lim_(x->c) (f(x))/(g(x)) = (lim_(x->c) f(x))/(lim_(x->c) g(x))` .

`= (lim_(n->oo)5)/(lim_(n->oo) (1+2/n^2))`

Take note that the limit of a constant is equal to itself `lim_(x->c) a = a.`

Also, if the rational function has a form `a/x^m` , where m represents any positive integer, its limit as x approaches infinity is zero `lim_(x->oo) (a/x^m) = 0` .

` (lim_(n->oo)5)/(lim_(n->oo) (1+2/n^2))`

`= 5/1`

`=5`

**Therefore, the limit of the given sequence is 5.**