1. **Problem statement:** Prove that $\lim_{t \to 0^+} f(t)$ exists for $f(t) = \frac{\sin 2t}{t}$ and find the Laplace transform of $f(t)$. Then use it to evaluate $\int_0^\infty \frac{e^{-3t} \sin 2t}{t} dt$. 2. **Step 1: Find the limit $\lim_{t \to 0^+} \frac{\sin 2t}{t}$.** Recall the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Rewrite: $$\lim_{t \to 0^+} \frac{\sin 2t}{t} = \lim_{t \to 0^+} 2 \cdot \frac{\sin 2t}{2t} = 2 \cdot \lim_{u \to 0} \frac{\sin u}{u} = 2 \cdot 1 = 2.$$ So the limit exists and equals 2. 3. **Step 2: Find the Laplace transform $\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt$ for $f(t) = \frac{\sin 2t}{t}$.** Recall the Laplace transform formula: $$\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} \frac{\sin 2t}{t} dt.$$ This integral is a known Laplace transform related to the sine integral function. Using the formula: $$\int_0^\infty \frac{\sin(at)}{t} e^{-st} dt = \arctan\left(\frac{a}{s}\right), \quad s > 0, a > 0.$$ Here, $a=2$, so: $$\mathcal{L}\left\{\frac{\sin 2t}{t}\right\} = \arctan\left(\frac{2}{s}\right).$$ 4. **Step 3: Use this to evaluate $\int_0^\infty \frac{e^{-3t} \sin 2t}{t} dt$.** Set $s=3$ in the Laplace transform: $$\int_0^\infty \frac{e^{-3t} \sin 2t}{t} dt = \arctan\left(\frac{2}{3}\right).$$ **Final answer:** $$\boxed{\arctan\left(\frac{2}{3}\right)}.$$