To evaluate the limit $\lim_{{x \to 0}} |x| \sin(x)$, we can analyze it by considering the properties of the absolute value and the sine function.

Step 1: Consider the behavior of $|x|$ and $\sin(x)$

• The function $|x|$ is the absolute value of $x$. It behaves as $x$ when $x$ is positive and as $-x$ when $x$ is negative.
• The sine function $\sin(x)$ is bounded between $-1$ and $1$ for all $x$.

Step 2: Establish bounds using the Squeeze Theorem

We can bound the product $|x| \sin(x)$ between $-|x|$ and $|x|$:

$-|x| \leq |x| \sin(x) \leq |x|$

Step 3: Apply the Squeeze Theorem

As $x \to 0$, both $-|x|$ and $|x|$ approach 0. Since $|x| \sin(x)$ is squeezed between these two functions, it must also approach 0.

Thus, by the Squeeze Theorem:

$\lim_{{x \to 0}} |x| \sin(x) = 0$

The final answer is:

$\lim_{{x \to 0}} |x| \sin(x) = 0$

Would you like any more detailed explanation on this, or do you have further questions?

Here are 5 related questions for further practice:

1. What is $\lim_{{x \to 0}} x \sin\left(\frac{1}{x}\right)$?
2. How would you evaluate $\lim_{{x \to 0}} x^2 \sin\left(\frac{1}{x}\right)$?
3. Can you find $\lim_{{x \to 0}} |x| \cos(x)$?
4. What is the limit $\lim_{{x \to 0}} x^2 \sin(x)$?
5. How do you apply the Squeeze Theorem to evaluate $\lim_{{x \to 0}} \frac{\sin(x)}{x}$?

Tip: The Squeeze Theorem is particularly useful when you have a product of a function that approaches 0 and a bounded function like sine or cosine.