To calculate the limit of the sequence given by

$x_n = \frac{n^2 + 3n}{n - 9} + \frac{4n^2 - n^3}{n^2 - 81},$

we can analyze each term separately as $n \to \infty$.

Step-by-Step Solution

1. First Term: $\frac{n^2 + 3n}{n - 9}$

   • As $n \to \infty$, the higher power of $n$ in the numerator and denominator dominates.
   • So, we simplify by dividing both the numerator and denominator by $n$: $\frac{n^2 + 3n}{n - 9} = \frac{n^2 \left(1 + \frac{3}{n}\right)}{n \left(1 - \frac{9}{n}\right)} = \frac{n \cdot (1 + \frac{3}{n})}{1 - \frac{9}{n}}.$
   • As $n \to \infty$, terms with $\frac{1}{n}$ tend to zero, leaving: $\frac{n \cdot 1}{1} = n.$

2. Second Term: $\frac{4n^2 - n^3}{n^2 - 81}$

   • Here, $n^3$ in the numerator and $n^2$ in the denominator are the dominant terms as $n \to \infty$.
   • Simplify by dividing the numerator and denominator by $n^2$: $\frac{4n^2 - n^3}{n^2 - 81} = \frac{n^2 \left(4 - \frac{n}{n^2}\right)}{n^2 \left(1 - \frac{81}{n^2}\right)} = \frac{4 - n}{1 - 0} = -n.$

3. Combine Results

   • The first term approaches $n$ and the second term approaches $-n$.
   • Therefore, as $n \to \infty$: $x_n = n + (-n) = 0.$

Final Answer

$\lim_{{n \to \infty}} x_n = 0.$

Would you like further details or have questions on any step?

Here are related questions to expand your understanding:

1. What happens if we modify the power of $n$ in the numerator of each fraction?
2. How do we evaluate limits with terms that grow at different rates?
3. Can the result change if we add a constant term to each fraction?
4. What other techniques could simplify such limits?
5. How does the limit change if we replace $n$ with $n^2$ in certain terms?

Tip: When finding limits at infinity, focus on the highest power terms, as they dominate the behavior of the function.