Originally Posted by

**MarkFL2** Once I have the form:

$\displaystyle A_n=\frac{(n+1)(11n+1)}{6n^2}$

I know it is of the form:

$\displaystyle A_n=\frac{11n^2+\cdots}{6n^2}$

and since the degree is the same in the numerator and denominator, I know the limit as $\displaystyle n\to\infty$ is simply the ratio of the leading coefficients, so we must have:

$\displaystyle \lim_{n\to\infty}A_n=\frac{11}{6}$

There is nothing wrong with your approach either, I just find it simpler to think of the technique we are taught in precalculus to find horizontal asymptotes.