The given problem is:

$\lim_{{x \to 0}} x \sin\left(\frac{1}{x}\right)$

Solution:

We need to evaluate the limit of $x \sin\left(\frac{1}{x}\right)$ as $x$ approaches 0. Here, we have a product of $x$ and $\sin\left(\frac{1}{x}\right)$, where $\sin\left(\frac{1}{x}\right)$ oscillates between -1 and 1 as $x \to 0$, but this doesn't directly give us the limit.

Step-by-step analysis:

• The sine function is bounded, i.e., for all real numbers $t$, $-1 \leq \sin(t) \leq 1$
• Thus, for $x \neq 0$, $-x \leq x \sin\left(\frac{1}{x}\right) \leq x$
• As $x \to 0$, both $-x$ and $x$ approach 0.

Using the Squeeze Theorem (or Sandwich Theorem), we conclude that: $\lim_{{x \to 0}} x \sin\left(\frac{1}{x}\right) = 0$

Final answer:

$\lim_{{x \to 0}} x \sin\left(\frac{1}{x}\right) = 0$

Would you like more details on this solution or have any further questions? Here are five related questions to think about:

1. How does the Squeeze Theorem work in general, and can it be applied to other trigonometric functions?
2. What is the behavior of $\sin\left(\frac{1}{x}\right)$ as $x \to 0$ by itself without multiplying by $x$?
3. Could L'Hopital's Rule be used in this situation? Why or why not?
4. How does this problem relate to the concept of oscillating functions in calculus?
5. What would happen if the limit was instead for $x \to \infty$?

Tip: The Squeeze Theorem is an extremely useful tool for handling limits involving oscillating functions, especially when those functions are bounded by simpler expressions!