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We are asked to evaluate the limit:

\[
\lim_{{x \to 0^+}} x \left( 1 - \cos \left( \frac{1}{x} \right) \right)
\]

using the **Squeeze Theorem**. Let's work through this step-by-step.

### Step 1: Analyze the function
The term we are interested in is:

\[
1 - \cos \left( \frac{1}{x} \right)
\]

We know that the cosine function oscillates between -1 and 1 for any real input, so:

\[
-1 \leq \cos \left( \frac{1}{x} \right) \leq 1
\]

From this, we subtract \( \cos \left( \frac{1}{x} \right) \) from 1:

\[
0 \leq 1 - \cos \left( \frac{1}{x} \right) \leq 2
\]

Thus, the term \( 1 - \cos \left( \frac{1}{x} \right) \) is bounded by 0 and 2:

\[
0 \leq 1 - \cos \left( \frac{1}{x} \right) \leq 2
\]

### Step 2: Multiply by \( x \)
Now we multiply the inequality by \( x \) (since \( x > 0 \) as \( x \to 0^+ \)):

\[
0 \leq x \left( 1 - \cos \left( \frac{1}{x} \right) \right) \leq 2x
\]

### Step 3: Apply the limit using the Squeeze Theorem
We now apply the limit to the inequality. As \( x \to 0^+ \):

\[
\lim_{{x \to 0^+}} 0 = 0
\quad \text{and} \quad
\lim_{{x \to 0^+}} 2x = 0
\]

By the **Squeeze Theorem**, since:

\[
0 \leq x \left( 1 - \cos \left( \frac{1}{x} \right) \right) \leq 2x
\]

and both the upper and lower bounds tend to 0 as \( x \to 0^+ \), we conclude:

\[
\lim_{{x \to 0^+}} x \left( 1 - \cos \left( \frac{1}{x} \right) \right) = 0
\]

### Final Answer:
\[
\lim_{{x \to 0^+}} x \left( 1 - \cos \left( \frac{1}{x} \right) \right) = 0
\]

Would you like more details on any part of this solution? Here are some related questions for further exploration:

1. How does the Squeeze Theorem work in general for oscillating functions?
2. What other types of limits can the Squeeze Theorem be used to solve?
3. Can you provide an example where the cosine function is used in a different type of limit?
4. How does the cosine function's behavior influence limits like this?
5. What are some examples of limits where the Squeeze Theorem cannot be used?

**Tip:** When dealing with oscillatory functions like sine or cosine, always consider their range to see if the Squeeze Theorem can help evaluate limits.

Use the Squeeze Theorem to evaluate \( \lim_{{x \to 0^+}} x \left( 1 - \cos \left( \frac{1}{x} \right) \right) \).

Learn how to evaluate the limit of x(1 - cos(1/x)) as x approaches 0 from the right using the Squeeze Theorem. This solution explores the behavior of oscillating functions and demonstrates the application of the Squeeze Theorem step-by-step, suitable for students in Grades 11-12.

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