1. The problem asks which limit expression equals the derivative of the function $f(x) = \sqrt{x}$ at $x=4$. 2. Recall the definition of the derivative at a point $a$: $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$ 3. For $f(x) = \sqrt{x}$ and $a=4$, this becomes: $$f'(4) = \lim_{x \to 4} \frac{\sqrt{x} - \sqrt{4}}{x - 4} = \lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$$ 4. Now, compare this with the given options: - A: $\lim_{x \to 4} \frac{\sqrt{x} - 4}{x - 4}$ (incorrect, subtracts 4 instead of 2) - B: $\lim_{x \to 4} \frac{\sqrt{4+x} - 2}{x - 4}$ (incorrect, inside root is $4+x$ not $x$) - C: $\lim_{x \to 4} \frac{\sqrt{x-4}}{x - 4}$ (incorrect, numerator is $\sqrt{x-4}$ not $\sqrt{x} - 2$) - D: $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$ (correct, matches the derivative definition) 5. Therefore, the correct answer is option D. 6. To confirm, the derivative of $f(x) = \sqrt{x}$ is: $$f'(x) = \frac{1}{2\sqrt{x}}$$ So, $$f'(4) = \frac{1}{2 \times 2} = \frac{1}{4}$$ 7. The limit in option D evaluates to $\frac{1}{4}$, confirming it represents $f'(4)$. Final answer: Option D