Determine whether the sequence converges or diverges. If it converges, find the limit. sqrt((2+9n^2)/(1+n^2))

A sequence {a_{n}} converges to a number L if for every positive number epsi there exists a number N such that |a_{n}-L|N .
(That is, for every index n>N, the value of the sequence is within an arbitrarily small number of the limit L.)
We are asked to determine if the sequence generated by a_{n}=sqrt{(2+9n^2)/(1+n^2)} converges, and if so to find the limit.
The sequence converges as it is bounded above. We can calculate the limit by rewriting the expression used to generate the elements of the sequence:
Rewrite sqrt{(2+9n^2)/(1+n^2)} by dividing the interior numerator and denominator by n^2 to get sqrt{(2/n^2+9)/(1/n^2+1)} . Now take the limit as n tends to infinity:
lim_(n->oo)sqrt{(2/n^2+9)/(1/n^2+1)}=sqrt(9)=3
(Note that we can rewrite the square root of a quotient as the quotient of square roots, and the limit of a quotient is the quotient of the respective limits. Then the limit as n tends to infinity of c/n^{x}=0 and the limit of a sum is the sum of the limits.)
The sequence converges to the limit 3.
Here is a graph of the continuous model for the sequence; note that for the graph of the sequence we should limit the domain to nonnegative integers.)
http://mathworld.wolfram.com/ConvergentSequence.html