To find the limit of the sequence

$a_n = \frac{9n^2 + 6n + 7}{8n^2 + 4n + 2}$

as $n \to \infty$, we analyze the terms by dividing both the numerator and denominator by $n^2$, the highest power of $n$ in the denominator:

$\lim_{n \to \infty} \frac{9n^2 + 6n + 7}{8n^2 + 4n + 2} = \lim_{n \to \infty} \frac{\frac{9n^2}{n^2} + \frac{6n}{n^2} + \frac{7}{n^2}}{\frac{8n^2}{n^2} + \frac{4n}{n^2} + \frac{2}{n^2}}.$

Simplify each term:

$\lim_{n \to \infty} \frac{9 + \frac{6}{n} + \frac{7}{n^2}}{8 + \frac{4}{n} + \frac{2}{n^2}}.$

As $n \to \infty$, the terms $\frac{6}{n}, \frac{7}{n^2}, \frac{4}{n},$ and $\frac{2}{n^2}$ approach 0. This simplifies the expression to:

$\frac{9}{8}.$

Thus, the limit is:

$\boxed{\frac{9}{8}}.$

Would you like a more detailed step-by-step breakdown or further clarification?

Related Questions:

1. How do you determine which term dominates as $n \to \infty$?
2. Why do we divide by the highest power of $n$ in the denominator?
3. What happens if the highest powers in the numerator and denominator are not equal?
4. How do limits behave when terms include logarithms or exponential growth?
5. How can this method be applied to similar rational functions?

Tip:

Always check for the highest degree term in both numerator and denominator when evaluating limits of rational functions.