1. **State the problem:** Find the derivative of the function $f(x) = 5 - x^3$ using the limit definition of the derivative. 2. **Recall the limit definition of the derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ This formula calculates the slope of the tangent line to the curve at any point $x$. 3. **Apply the definition to $f(x) = 5 - x^3$:** Calculate $f(x+h)$: $$f(x+h) = 5 - (x+h)^3$$ 4. **Expand $(x+h)^3$ using the binomial theorem:** $$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$$ 5. **Substitute back into the difference quotient:** $$\frac{f(x+h) - f(x)}{h} = \frac{[5 - (x^3 + 3x^2h + 3xh^2 + h^3)] - (5 - x^3)}{h}$$ 6. **Simplify the numerator:** $$= \frac{5 - x^3 - 3x^2h - 3xh^2 - h^3 - 5 + x^3}{h} = \frac{-3x^2h - 3xh^2 - h^3}{h}$$ 7. **Factor out $h$ in the numerator and cancel:** $$= \frac{h(-3x^2 - 3xh - h^2)}{h} = -3x^2 - 3xh - h^2$$ 8. **Take the limit as $h \to 0$:** $$f'(x) = \lim_{h \to 0} (-3x^2 - 3xh - h^2) = -3x^2$$ **Final answer:** $$\boxed{f'(x) = -3x^2}$$ This means the slope of the tangent line to the curve $f(x) = 5 - x^3$ at any point $x$ is $-3x^2$.