To evaluate the limit:

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

we can use L'Hôpital's Rule. However, let's first rewrite the expression to determine if it's an indeterminate form:

1. Rewrite $x \sin\left(\frac{1}{x}\right)$ as: $\lim_{x \to \infty} \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$

2. Now, as $x \to \infty$, $\frac{1}{x} \to 0$, and since $\sin\left(\frac{1}{x}\right) \to 0$ as well, we get a $\frac{0}{0}$ indeterminate form, which allows us to apply L'Hôpital's Rule.

3. Differentiate the numerator and the denominator with respect to $x$:

   • The derivative of the numerator, $\sin\left(\frac{1}{x}\right)$, using the chain rule, is: $\cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = -\frac{\cos\left(\frac{1}{x}\right)}{x^2}$
   • The derivative of the denominator, $\frac{1}{x}$, is: $-\frac{1}{x^2}$

4. Substitute these derivatives into the limit: $\lim_{x \to \infty} \frac{-\frac{\cos\left(\frac{1}{x}\right)}{x^2}}{-\frac{1}{x^2}} = \lim_{x \to \infty} \cos\left(\frac{1}{x}\right)$

5. As $x \to \infty$, $\frac{1}{x} \to 0$, so $\cos\left(\frac{1}{x}\right) \to \cos(0) = 1$.

Therefore:

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

The answer is $1$.

Would you like a detailed explanation of each step?

Here are some related questions to deepen your understanding:

1. How does L'Hôpital's Rule help with evaluating indeterminate forms?
2. What are some other common indeterminate forms where L'Hôpital's Rule is useful?
3. How would the solution change if we had a different trigonometric function, like cosine, instead of sine?
4. Can we solve this limit without using L'Hôpital's Rule?
5. What does it mean when we say that a limit "tends to" a particular value?

Tip: When using L'Hôpital's Rule, always confirm that the original expression is indeed an indeterminate form, like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.