1. **Problem:** Evaluate the limit $$\lim_{t \to \infty} \frac{\ln(3t)}{t^2}$$ using L'Hospital's Rule. 2. **Formula and rule:** L'Hospital's Rule states that if $$\lim_{x \to a} \frac{f(x)}{g(x)}$$ results in an indeterminate form $$\frac{\infty}{\infty}$$ or $$\frac{0}{0}$$, then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ provided the latter limit exists. 3. **Check the form:** As $$t \to \infty$$, $$\ln(3t) \to \infty$$ and $$t^2 \to \infty$$, so the form is $$\frac{\infty}{\infty}$$, suitable for L'Hospital's Rule. 4. **Differentiate numerator and denominator:** $$f(t) = \ln(3t) \implies f'(t) = \frac{1}{3t} \cdot 3 = \frac{1}{t}$$ $$g(t) = t^2 \implies g'(t) = 2t$$ 5. **Apply L'Hospital's Rule:** $$\lim_{t \to \infty} \frac{\ln(3t)}{t^2} = \lim_{t \to \infty} \frac{\frac{1}{t}}{2t} = \lim_{t \to \infty} \frac{1}{2t^2}$$ 6. **Evaluate the new limit:** As $$t \to \infty$$, $$\frac{1}{2t^2} \to 0$$. 7. **Final answer:** $$\boxed{0}$$