1. **State the problem:** Find the limit $$\lim_{x \to \pi} (x - \pi) \cos^2 \left(\frac{1}{x - \pi}\right)$$ 2. **Recall the formula and important rules:** We know that if $f(x)$ is bounded and $g(x) \to 0$, then $g(x)f(x) \to 0$. Here, $\cos^2(\cdot)$ is bounded between 0 and 1. 3. **Analyze the components:** - As $x \to \pi$, the term $(x - \pi) \to 0$. - The term $\cos^2\left(\frac{1}{x - \pi}\right)$ oscillates between 0 and 1 infinitely often because $\frac{1}{x - \pi}$ goes to infinity. 4. **Use the squeeze theorem:** Since $0 \leq \cos^2\left(\frac{1}{x - \pi}\right) \leq 1$, multiply all parts by $|x - \pi|$ (which tends to 0): $$0 \leq |(x - \pi) \cos^2\left(\frac{1}{x - \pi}\right)| \leq |x - \pi|$$ 5. **Apply the limit:** $$\lim_{x \to \pi} 0 = 0$$ $$\lim_{x \to \pi} |x - \pi| = 0$$ By the squeeze theorem, $$\lim_{x \to \pi} (x - \pi) \cos^2 \left(\frac{1}{x - \pi}\right) = 0$$ **Final answer:** $$\boxed{0}$$