https://file.aiquickdraw.com/mathbot/file/1725402505232u38rarit.jpg

The image contains a mathematical expression that appears to be a limit problem:

\[
\lim_{x \to \infty} \left( \frac{\sin(2x)}{3x} \right)
\]

### Solving the Limit:

To solve this limit, let's analyze the expression:

\[
\lim_{x \to \infty} \left( \frac{\sin(2x)}{3x} \right)
\]

1. **Bounded Nature of Sine Function**:  
   The sine function, \(\sin(2x)\), is bounded between -1 and 1 for all values of \(x\):
   \[
   -1 \leq \sin(2x) \leq 1
   \]

2. **Considering the Entire Expression**:  
   When dividing the sine function by \(3x\), as \(x\) approaches infinity, the denominator \(3x\) grows without bounds. Therefore, the fraction \(\frac{\sin(2x)}{3x}\) becomes smaller and smaller in magnitude.

3. **Squeeze Theorem Application**:
   Since \(-\frac{1}{3x} \leq \frac{\sin(2x)}{3x} \leq \frac{1}{3x}\), and both \(\frac{1}{3x}\) and \(-\frac{1}{3x}\) tend to 0 as \(x \to \infty\), by the Squeeze Theorem, the limit of the original expression is also 0.

Thus, the limit is:

\[
\boxed{0}
\]

Would you like more details on this solution or have any other questions?

### Relative Questions:
1. What is the Squeeze The

What is the limit of (sin(2x))/(3x) as x approaches infinity?

Understand how to calculate the limit of (sin(2x))/(3x) as x approaches infinity using the Squeeze Theorem. This problem covers advanced calculus concepts and is suitable for college-level students.

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