Question

use the appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges. (if the quantity diverges, enter diverg)

\\a_n = \frac{9n^2 + n + 4}{4n^2 - 9}\\

\\\lim_{n \to \infty} a_n = \\

Explanation:

Divide numerator and denominator by the highest power of n

$$ a_n = \frac{\frac{9n^2}{n^2} + \frac{n}{n^2} + \frac{4}{n^2}}{\frac{4n^2}{n^2} - \frac{9}{n^2}} = \frac{9 + \frac{1}{n} + \frac{4}{n^2}}{4 - \frac{9}{n^2}} $$

Apply limit laws for sequences

$$ \lim_{n \to \infty} a_n = \frac{\lim_{n \to \infty} 9 + \lim_{n \to \infty} \frac{1}{n} + \lim_{n \to \infty} \frac{4}{n^2}}{\lim_{n \to \infty} 4 - \lim_{n \to \infty} \frac{9}{n^2}} $$

Evaluate the limits of individual terms

$$ \lim_{n \to \infty} a_n = \frac{9 + 0 + 0}{4 - 0} = \frac{9}{4} $$

Answer:

Use the appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges. (If the quantity diverges, enter DIVERGES)
\(a_n = \frac{9n^2 + n + 4}{4n^2 - 9}\)
\(\lim_{n \to \infty} a_n =\) <blank>\(\frac{9}{4}\)</blank>