1. **State the problem:** We need to find the value of the limit $$\lim_{x \to 4} \frac{f(x)}{x^2 - 16}$$ where the function $f(x)$ is given by a graph consisting of semicircles and line segments. 2. **Recall the formula and important rules:** The denominator can be factored as $$x^2 - 16 = (x-4)(x+4)$$. 3. **Evaluate the denominator at $x=4$:** $$4^2 - 16 = 16 - 16 = 0$$ So the denominator approaches zero as $x \to 4$. 4. **Evaluate $f(4)$ from the graph:** The graph shows a line segment from $(2,2)$ to $(5,-1)$. We find $f(4)$ by linear interpolation: Slope of segment: $$m = \frac{-1 - 2}{5 - 2} = \frac{-3}{3} = -1$$ Equation of line segment: $$y - 2 = -1(x - 2) \Rightarrow y = -x + 4$$ At $x=4$: $$f(4) = -4 + 4 = 0$$ 5. **Since both numerator and denominator approach zero, apply L'Hôpital's Rule:** We need $f'(4)$ and derivative of denominator: Denominator derivative: $$\frac{d}{dx}(x^2 - 16) = 2x$$ At $x=4$: $$2 \times 4 = 8$$ 6. **Find $f'(4)$:** Since $f(x)$ is linear between $(2,2)$ and $(5,-1)$, the derivative is the slope: $$f'(4) = -1$$ 7. **Apply L'Hôpital's Rule:** $$\lim_{x \to 4} \frac{f(x)}{x^2 - 16} = \lim_{x \to 4} \frac{f'(x)}{2x} = \frac{f'(4)}{8} = \frac{-1}{8}$$ **Final answer:** $$\boxed{-\frac{1}{8}}$$ This corresponds to option C.