1. **Problem:** Find the limit $\lim_{x \to 0} x \cot x$. 2. **Formula and rules:** Recall that $\cot x = \frac{\cos x}{\sin x}$ and near zero, $\sin x \approx x$ and $\cos x \approx 1$. 3. **Work:** $$x \cot x = x \cdot \frac{\cos x}{\sin x} = \frac{x \cos x}{\sin x}$$ As $x \to 0$, $\sin x \approx x$, so $$\lim_{x \to 0} \frac{x \cos x}{\sin x} = \lim_{x \to 0} \frac{x \cos x}{x} = \lim_{x \to 0} \cos x = 1$$ 4. **Answer:** $$\boxed{1}$$