1. The problem is to find the limit: $$\lim_{x \to 2} \frac{3}{x-3} + 1$$. 2. Recall that the limit of a sum is the sum of the limits, provided the limits exist. 3. We can write: $$\lim_{x \to 2} \frac{3}{x-3} + 1 = \lim_{x \to 2} \frac{3}{x-3} + \lim_{x \to 2} 1$$. 4. The limit of a constant is the constant itself, so: $$\lim_{x \to 2} 1 = 1$$. 5. Now evaluate $$\lim_{x \to 2} \frac{3}{x-3}$$. 6. Substitute $x=2$ directly: $$\frac{3}{2-3} = \frac{3}{-1} = -3$$. 7. Therefore, the limit is: $$-3 + 1 = -2$$. Final answer: $$\boxed{-2}$$