To find the limit of the sequence $u_n = \frac{3n + 4}{3 \cos(n) + 2n^2}$ as $n$ approaches infinity, we analyze the behavior of the numerator and denominator as $n$ grows large.

First, observe the leading terms in both the numerator and the denominator:

• The numerator: $3n + 4$
• The denominator: $3 \cos(n) + 2n^2$

For large $n$, the term $3n$ dominates in the numerator, and $2n^2$ dominates in the denominator because $n^2$ grows faster than any linear term or the bounded $\cos(n)$.

We can approximate the sequence for large $n$ by ignoring the less significant terms:

$u_n \approx \frac{3n}{2n^2} = \frac{3}{2n}$

Now, as $n$ approaches infinity, the term $\frac{3}{2n}$ approaches 0.

Therefore, the limit of the sequence $u_n$ is:

$\lim_{n \to \infty} u_n = 0$

So, the limit of the sequence $(u_n)$ is $0$.

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How do you determine the dominant terms in a sequence or function as $n$ approaches infinity?
2. What is the significance of bounded functions like $\cos(n)$ when analyzing limits?
3. How do you apply L'Hôpital's rule to sequences?
4. What other methods can be used to find limits of sequences?
5. Can you explain how polynomial terms grow relative to each other?

Tip: When analyzing limits, always focus on the terms that grow the fastest as $n$ approaches infinity. These dominant terms typically dictate the behavior of the sequence or function.