The given expression is:

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

Solution:

1. As $x \to \infty$, notice that $\frac{2}{x} \to 0$.

2. We know that for small values of $y$, $\sin(y) \approx y$. Thus, when $\frac{2}{x}$ becomes very small, we can approximate: $\sin\left(\frac{2}{x}\right) \approx \frac{2}{x}.$

3. Applying this approximation: $\sin\left(\sin\left(\frac{2}{x}\right)\right) \approx \sin\left(\frac{2}{x}\right) \approx \frac{2}{x}.$

4. Now substitute this approximation into the original limit: $\lim_{x \to \infty} x \cdot \sin\left(\sin\left(\frac{2}{x}\right)\right) \approx \lim_{x \to \infty} x \cdot \frac{2}{x}.$

5. This simplifies to: $\lim_{x \to \infty} 2 = 2.$

Conclusion:

The value of the limit is $\boxed{2}$.

Would you like further clarification or details on any part of the solution?

Here are 5 follow-up questions you might find useful:

1. What is the small angle approximation for trigonometric functions?
2. How can we justify using approximations for limits as $x \to \infty$?
3. What happens if we use higher-order approximations for the sine function in this problem?
4. Can you explain how limits behave for nested functions like $\sin(\sin(x))$?
5. How does L'Hôpital's Rule apply to cases like this?

Tip: In limits involving trigonometric functions, using approximations for small angles can greatly simplify the calculations.