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 $x \to 0^+$: