1. **Problem:** Find the limit $$\lim_{t \to 0} t \tan t$$ using L'Hospital's rule. 2. **Check form:** As $t \to 0$, $t \to 0$ and $\tan t \to 0$, so the product $t \tan t \to 0$. This is an indeterminate form of type $0 \cdot 0$, which can be rewritten as a quotient to apply L'Hospital's rule. 3. **Rewrite:** $$t \tan t = \frac{\tan t}{\frac{1}{t}}$$ 4. **Apply L'Hospital's rule:** Differentiate numerator and denominator with respect to $t$: $$\frac{d}{dt}(\tan t) = \sec^2 t$$ $$\frac{d}{dt}\left(\frac{1}{t}\right) = -\frac{1}{t^2}$$ 5. **Evaluate new limit:** $$\lim_{t \to 0} \frac{\sec^2 t}{-\frac{1}{t^2}} = \lim_{t \to 0} -t^2 \sec^2 t$$ 6. **Simplify:** As $t \to 0$, $\sec^2 t \to 1$, so $$\lim_{t \to 0} -t^2 \cdot 1 = 0$$ **Final answer:** $$\boxed{0}$$