The answer for the limit in question is:

$\lim_{x \to \infty} x \sin\left(\frac{1}{x}\right) = 1$

This conclusion follows from the steps shown in the solution. The key part is using the substitution $y = \frac{1}{x}$, which converts the limit into a known standard limit:

$\lim_{y \to 0} \frac{\sin(y)}{y} = 1$

Thus, the original limit evaluates to 1.

Would you like any further clarification on this problem?

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Here are some more questions related to this problem:

1. How does substituting $y = \frac{1}{x}$ simplify the problem?
2. Can the result be generalized for other trigonometric functions?
3. What would happen if the function inside the sine were different, e.g., $\sin\left(\frac{2}{x}\right)$?
4. What are the conditions under which the limit $\lim_{y \to 0} \frac{\sin(y)}{y} = 1$ holds true?
5. How can we verify the result using L'Hôpital's Rule?
6. What if the limit was $\lim_{x \to 0^+} x \sin\left(\frac{1}{x}\right)$?
7. Can this approach be applied to other types of limits involving trigonometric functions?
8. What is the geometric interpretation of $\lim_{y \to 0} \frac{\sin(y)}{y}$?

Tip: Always check whether a limit problem can be simplified by a substitution, especially when dealing with trigonometric functions or other complex expressions.