Let's analyze the problem and its solution from the images provided.

Problem Statement:

The task is to find the limit $\lim_{x \to \infty} x \sin\left(\frac{1}{x}\right)$.

Steps for the Solution:

1. Substitution:

   • Let $y = \frac{1}{x}$.
   • As $x \to \infty$, $y \to 0$.

2. Rewriting the Limit:

   • The original limit is rewritten using the substitution: $\lim_{x \to \infty} x \sin\left(\frac{1}{x}\right)$ becomes $\lim_{y \to 0} \frac{\sin(y)}{y}$.

3. Using the Known Limit:

   • We use the standard limit result $\lim_{y \to 0} \frac{\sin(y)}{y} = 1$.

4. Final Conclusion:

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

The solution in the images accurately guides through these steps to arrive at the correct limit.

Would you like any further details or explanations?

Here are some related questions you might consider:

1. What is the general form of a limit involving trigonometric functions?
2. How would the solution differ if we were to consider $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$?
3. Can the L'Hôpital's rule be applied to this limit problem? If so, how?
4. What other standard limits should one memorize for similar problems?
5. How does the substitution method simplify the evaluation of complex limits?
6. How can we intuitively understand the behavior of $x \sin\left(\frac{1}{x}\right)$ as $x \to \infty$?
7. What are some potential pitfalls when evaluating trigonometric limits at infinity?
8. Could this problem be extended to evaluate $\lim_{x \to \infty} x \sin\left(\frac{1}{x^n}\right)$ for any $n$?

Tip: Memorizing standard trigonometric limits like $\lim_{y \to 0} \frac{\sin(y)}{y} = 1$ can significantly simplify solving more complex limit problems.