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