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.

Learn how to evaluate the limit of the expression (2/x * sin(x)) as x approaches infinity. This problem involves understanding limits, oscillating functions like sine, and applying the Squeeze Theorem. Suitable for high school or early college-level students.

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