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Let's analyze and solve the given limit:

\[
\lim_{n \to \infty} \left[ \left( \frac{n^4 + 2n^2}{n^4 + n^2 + 1} \right)^{3n^2} + \frac{\sin(n^4 + 2n^2)}{3n^2} \right]
\]

### Step-by-step analysis:

1. **Analyzing the first part:**

   \[
   \left( \frac{n^4 + 2n^2}{n^4 + n^2 + 1} \right)^{3n^2}
   \]

   - As \(n \to \infty\), the highest degree terms dominate in the numerator and denominator. Both the numerator and the denominator have the term \(n^4\), so:

     \[
     \frac{n^4 + 2n^2}{n^4 + n^2 + 1} \to 1
     \]

   - This means that the base of the exponential expression approaches 1.

   - Now, we analyze the limit of the exponent:

     \[
     \left( 1 \right)^{3n^2} = 1
     \]

   Therefore, the first part of the expression tends to \(1\) as \(n \to \infty\).

2. **Analyzing the second part:**

   \[
   \frac{\sin(n^4 + 2n^2)}{3n^2}
   \]

   - The sine function oscillates between -1 and 1, and as \(n \to \infty\), the term \(n^4 + 2n^2\) grows very large.
   
   - However, since sine is bounded, we know:

     \[
     \frac{\sin(n^4 + 2n^2)}{3n^2} \to 0
     \]

     because the denominator \(3n^2\) grows without bound.

### Final result:

The limit of the given expression as \(n \to \infty\) is:

\[
\lim_{n \to \infty} \left( 1 + 0 \right) = 1
\]

Thus, the limit exists and is equal to 1.

---

Would you like more details on any step or further clarification?

Here are five follow-up questions based on this explanation:

1. What happens if we change the power of \(3n^2\) to a different function of \(n\)?
2. How does the behavior of the sine function influence the result?
3. What would the result be if the denominator in the sine term was \(n^4\) instead of \(3n^2\)?
4. Can L'Hopital's rule be applied to simplify the sine term further?
5. How does the highest degree term in the polynomial affect the behavior of the limit?

**Tip:** When dealing with limits involving polynomials, always focus on the terms with the highest degree, as they dominate the behavior of the expression as \(n \to \infty\).

Calculate, if it exists, the following limit: 
lim(n→∞) [( (n^4 + 2n^2) / (n^4 + n^2 + 1) )^(3n^2) + sin(n^4 + 2n^2) / (3n^2) ]

Explore the solution to the limit problem involving exponential and trigonometric functions. The problem examines the behavior of complex expressions as n approaches infinity, using key calculus concepts such as the Squeeze Theorem and properties of limits.

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