The given expression is:

$a_n = \frac{n^3 + 3n}{n^2 + 2}$

We are tasked with finding the limit of $a_n$ as $n \to \infty$.

Solution:

1. First, let's divide both the numerator and the denominator by $n^2$, the highest power of $n$ in the denominator.

$a_n = \frac{\frac{n^3 + 3n}{n^2}}{\frac{n^2 + 2}{n^2}} = \frac{\frac{n^3}{n^2} + \frac{3n}{n^2}}{\frac{n^2}{n^2} + \frac{2}{n^2}}$

2. Simplifying the terms:

$a_n = \frac{n + \frac{3}{n}}{1 + \frac{2}{n^2}}$

3. Now, as $n \to \infty$:

   • $\frac{3}{n} \to 0$
   • $\frac{2}{n^2} \to 0$

4. Therefore, the limit of $a_n$ as $n \to \infty$ is:

$\lim_{n \to \infty} a_n = \frac{n + 0}{1 + 0} = n$

So, the expression grows without bound as $n \to \infty$, which implies that:

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

Thus, the limit does not exist in a finite form since it approaches infinity.

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

Here are 5 related questions to consider:

1. How does dividing by the highest power of $n$ help in evaluating limits?
2. What would happen if the degrees of the numerator and denominator were different?
3. Can this method be used for non-polynomial expressions?
4. How does the growth rate of functions affect limits at infinity?
5. What happens to rational functions as $n \to \infty$ if the degree of the denominator is higher?

Tip: When evaluating limits of rational functions, simplifying by the highest power of $n$ helps highlight which terms dominate as $n \to \infty$.