To evaluate the limit

$\lim_{x \to \infty} \frac{\sin(x)}{x(x^2 - 1)},$

we proceed as follows:

Step 1: Analyze the behavior of the numerator and denominator

1. Numerator: The term $\sin(x)$ oscillates between $-1$ and $1$ for all $x$.
2. Denominator: The denominator $x(x^2 - 1) = x^3 - x$ grows without bound as $x \to \infty$, dominated by the $x^3$ term.

Step 2: Bound the fraction

Since $-1 \leq \sin(x) \leq 1$, we know $-\frac{1}{x(x^2 - 1)} \leq \frac{\sin(x)}{x(x^2 - 1)} \leq \frac{1}{x(x^2 - 1)}.$

Step 3: Evaluate the bounds

The bounding term $\frac{1}{x(x^2 - 1)}$ simplifies approximately to $\frac{1}{x^3}$ for large $x$. As $x \to \infty$, we have: $\frac{1}{x^3} \to 0.$ Thus, by the Squeeze Theorem: $\lim_{x \to \infty} \frac{\sin(x)}{x(x^2 - 1)} = 0.$

Final Answer:

$\boxed{0}$

Would you like a detailed explanation of the Squeeze Theorem or any step here?

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Expanding Thoughts:

1. What is the importance of the oscillatory nature of $\sin(x)$ in limits?
2. How does the growth rate of $x^3$ compare to other polynomial terms?
3. Could this limit change if $\sin(x)$ were replaced with $\cos(x)$?
4. What would happen if $x$ approached a finite value instead of infinity?
5. How can we generalize this approach to similar problems involving oscillatory functions and polynomials?

Tip: The Squeeze Theorem is powerful for handling limits with oscillatory functions if you can bound them between two converging terms.