1. **Problem:** Calculate the limit $$\lim_{x \to 0} \frac{3e^x - 3}{2x}$$. 2. **Formula and rules:** This is a limit of the form $$\frac{f(x) - f(a)}{x - a}$$ as $$x \to a$$, which resembles the definition of the derivative: $$\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$$. 3. **Identify function:** Here, $$f(x) = 3e^x$$ and $$a = 0$$. 4. **Calculate derivative:** $$f'(x) = 3e^x$$. 5. **Evaluate derivative at $$x=0$$:** $$f'(0) = 3e^0 = 3 \times 1 = 3$$. 6. **Therefore, the limit is:** $$\lim_{x \to 0} \frac{3e^x - 3}{2x} = \frac{1}{2} \times \lim_{x \to 0} \frac{3e^x - 3}{x} = \frac{1}{2} \times 3 = \frac{3}{2}$$. 7. **Intermediate step showing cancellation:** $$\lim_{x \to 0} \frac{3e^x - 3}{2x} = \lim_{x \to 0} \frac{3(e^x - 1)}{2x} = \frac{3}{2} \lim_{x \to 0} \frac{e^x - 1}{x}$$ Since $$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$, the limit is $$\frac{3}{2}$$. **Final answer:** $$\boxed{\frac{3}{2}}$$