We are tasked with finding the term independent of \( x \) in the expansion of:
\[
\left[ x + \frac{\sin(1/n)}{x^2} \right]^{3n}.
\]
This involves using the binomial expansion and focusing on the term that is independent of \( x \).
Step 1: Approximate \( \sin(1/n) \) for large \( n \).
Since \( \sin(1/n) \approx \frac{1}{n} \) for large \( n \), we approximate:
\[
\frac{\sin(1/n)}{x^2} \approx \frac{1}{n x^2}.
\]
Thus, the expression becomes:
\[
\left[ x + \frac{1}{n x^2} \right]^{3n}.
\]
Step 2: Binomial Expansion.
The binomial expansion of \( \left( x + \frac{1}{n x^2} \right)^{3n} \) is given by:
\[
\left( x + \frac{1}{n x^2} \right)^{3n} = \sum_{k=0}^{3n} \binom{3n}{k} x^{3n-k} \left( \frac{1}{n x^2} \right)^k.
\]
Simplifying each term:
\[
= \sum_{k=0}^{3n} \binom{3n}{k} \frac{x^{3n-k}}{n^k x^{2k}} = \sum_{k=0}^{3n} \binom{3n}{k} \frac{1}{n^k} x^{3n-3k-2k}.
\]
This simplifies to:
\[
= \sum_{k=0}^{3n} \binom{3n}{k} \frac{1}{n^k} x^{3n-5k}.
\]
Step 3: Finding the Independent Term.
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[
3n - 5k = 0 \quad \Rightarrow \quad k = \frac{3n}{5}.
\]
Since \( k \) must be an integer, \( n \) must be a multiple of 5 for the term independent of \( x \) to exist.
Step 4: Asymptotic Behavior.
As \( n \) becomes very large, the coefficient of the term independent of \( x \) behaves in such a way that:
\[
\lim_{n \to \infty} \frac{a_n n!}{3^n n^n} = 0.
\]
This is because the factorial growth of \( n! \) is not sufficient to offset the exponential decay due to the powers of \( n^n \) and \( 3^n \) in the denominator.
Step 5: Conclusion.
The value of the limit is \( 0 \).