1. **State the problem:** We are given a piecewise function $$f(x) = \begin{cases} -x^2 + 3x + 3 & \text{if } x < 2 \\ 6 & \text{if } x = 2 \\ 8 - \frac{3}{2}x & \text{if } x > 2 \end{cases}$$ We want to find the value of $$\lim_{x \to 2} f(f(x))$$. 2. **Understand the problem:** To find $$\lim_{x \to 2} f(f(x))$$, we first need to find $$\lim_{x \to 2} f(x)$$ and then evaluate $$f$$ at that limit if possible. 3. **Find $$\lim_{x \to 2^-} f(x)$$:** For $$x < 2$$, $$f(x) = -x^2 + 3x + 3$$ Evaluate the limit as $$x$$ approaches 2 from the left: $$\lim_{x \to 2^-} f(x) = -(2)^2 + 3(2) + 3 = -4 + 6 + 3 = 5$$ 4. **Find $$\lim_{x \to 2^+} f(x)$$:** For $$x > 2$$, $$f(x) = 8 - \frac{3}{2}x$$ Evaluate the limit as $$x$$ approaches 2 from the right: $$\lim_{x \to 2^+} f(x) = 8 - \frac{3}{2} \times 2 = 8 - 3 = 5$$ 5. **Check if $$\lim_{x \to 2} f(x)$$ exists:** Since the left and right limits are equal, $$\lim_{x \to 2} f(x) = 5$$ 6. **Evaluate $$f(f(x))$$ limit:** We want $$\lim_{x \to 2} f(f(x)) = \lim_{x \to 2} f(y) \text{ where } y = f(x)$$ Since $$\lim_{x \to 2} f(x) = 5$$, this becomes $$\lim_{y \to 5} f(y)$$ 7. **Determine which piece of $$f(y)$$ to use at $$y=5$$:** Since 5 is greater than 2, we use the piece for $$y > 2$$: $$f(y) = 8 - \frac{3}{2} y$$ 8. **Calculate $$f(5)$$:** $$f(5) = 8 - \frac{3}{2} \times 5 = 8 - 7.5 = 0.5$$ 9. **Final answer:** $$\lim_{x \to 2} f(f(x)) = 0.5$$