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The given problem is to compute the limit:

\[
\lim_{x \to \infty} \frac{2\sin{x} - 3\cos{x}}{x}
\]

### Step-by-step solution:

1. **Analyze the function**: The numerator consists of trigonometric functions \( 2\sin{x} \) and \( -3\cos{x} \), while the denominator is \( x \).
2. **Behavior of trigonometric functions**: Both \( \sin{x} \) and \( \cos{x} \) are bounded functions. Specifically, their values lie within the range \([-1, 1]\).
3. **Bound the numerator**: Since \( \sin{x} \) and \( \cos{x} \) are bounded, the entire numerator \( 2\sin{x} - 3\cos{x} \) is also bounded. We know:
   \[
   -2 \leq 2\sin{x} \leq 2
   \quad \text{and} \quad
   -3 \leq -3\cos{x} \leq 3
   \]
   Therefore, the numerator lies in the range:
   \[
   -5 \leq 2\sin{x} - 3\cos{x} \leq 5
   \]
4. **Divide by \( x \)**: As \( x \to \infty \), the denominator \( x \) grows without bound, but the numerator remains bounded. Hence, the fraction tends to 0 as \( x \to \infty \).

### Conclusion:
\[
\lim_{x \to \infty} \frac{2\sin{x} - 3\cos{x}}{x} = 0
\]

Would you like more details or clarifications?

Here are 5 related questions to explore:
1. What is the general behavior of trigonometric limits as \( x \to \infty \)?
2. How does L'Hôpital's rule apply to limits involving infinity?
3. What are other methods to compute limits of functions with oscillatory terms?
4. How does the behavior of bounded functions affect limits?
5. Can you compute similar limits for other trigonometric expressions?

**Tip**: When dealing with limits involving trigonometric functions and infinity, consider the bounded nature of sine and cosine functions to simplify your work.

Compute the limit: lim (x → ∞) [(2sinx - 3cosx) / x]

Learn how to compute the limit of (2sinx - 3cosx) / x as x approaches infinity. This problem involves trigonometric functions and their behavior in limits, showcasing how bounded functions impact the overall limit.

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