To evaluate the limit:

$\lim_{{x \to \infty}} \left( \frac{2}{x} \sin x \right)$

We can approach this by analyzing each part of the expression:

1. As $x \to \infty$, $\frac{2}{x}$ approaches 0, because the denominator grows larger without bound.
2. The function $\sin x$ oscillates between -1 and 1 for all values of $x$, so it does not tend to a specific limit.

Therefore, we have a product of something approaching 0 and something bounded. The boundedness of $\sin x$ means that $\frac{2}{x} \sin x$ is squeezed to 0 as $x \to \infty$.

Thus, the limit is:

$\lim_{{x \to \infty}} \left( \frac{2}{x} \sin x \right) = 0$

Would you like further clarification on any part of this? Here are some related questions:

1. What is the Squeeze Theorem, and how is it applied in limits?
2. How do oscillating functions like sine affect the behavior of limits?
3. Can the limit of $\frac{\sin x}{x}$ as $x \to \infty$ be generalized?
4. What are common limits involving trigonometric functions?
5. How do you evaluate limits involving products of small and oscillating functions?

Tip: When evaluating limits with oscillating functions, consider how boundedness and multiplication by vanishing factors (like $\frac{1}{x}$) affect the result.