1. **State the problem:** We want to find the limit $$\lim_{x \to 0} x \sqrt{\cos \sqrt{x}}.$$\n\n2. **Recall the limit and approximation rules:** As $x \to 0$, $\sqrt{x} \to 0$. We can use the approximation for cosine near zero: $$\cos y \approx 1 - \frac{y^2}{2}$$ for small $y$.\n\n3. **Apply the approximation:** Let $y = \sqrt{x}$. Then $$\cos \sqrt{x} = \cos y \approx 1 - \frac{y^2}{2} = 1 - \frac{x}{2}.$$\n\n4. **Evaluate the square root:** $$\sqrt{\cos \sqrt{x}} \approx \sqrt{1 - \frac{x}{2}}.$$ For small $x$, use the binomial approximation: $$\sqrt{1 - z} \approx 1 - \frac{z}{2}$$ for small $z$. Here, $z = \frac{x}{2}$. So, $$\sqrt{\cos \sqrt{x}} \approx 1 - \frac{x}{4}.$$\n\n5. **Multiply by $x$:** $$x \sqrt{\cos \sqrt{x}} \approx x \left(1 - \frac{x}{4}\right) = x - \frac{x^2}{4}.$$\n\n6. **Take the limit as $x \to 0$:** Both $x$ and $x^2$ go to zero, so $$\lim_{x \to 0} x \sqrt{\cos \sqrt{x}} = 0.$$\n\n**Final answer:** $$\boxed{0}.$$