1. **Problem:** Estimate the limit of the function $f(x) = \frac{x^3 - 125}{x - 5}$ as $x$ approaches 5. 2. **Formula and Important Rule:** The limit of a function as $x$ approaches a value $a$ is the value that $f(x)$ approaches as $x$ gets closer to $a$. If direct substitution results in an indeterminate form like $\frac{0}{0}$, we simplify the expression. 3. **Step 1: Direct substitution** Substitute $x=5$ into $f(x)$: $$f(5) = \frac{5^3 - 125}{5 - 5} = \frac{125 - 125}{0} = \frac{0}{0}$$ This is an indeterminate form, so we simplify. 4. **Step 2: Factor numerator** Note that $x^3 - 125$ is a difference of cubes: $$x^3 - 125 = (x - 5)(x^2 + 5x + 25)$$ 5. **Step 3: Simplify the function** $$f(x) = \frac{(x - 5)(x^2 + 5x + 25)}{x - 5}$$ Cancel the common factor $x - 5$: $$f(x) = \frac{\cancel{(x - 5)}(x^2 + 5x + 25)}{\cancel{(x - 5)}} = x^2 + 5x + 25$$ 6. **Step 4: Evaluate the simplified function at $x=5$** $$f(5) = 5^2 + 5(5) + 25 = 25 + 25 + 25 = 75$$ 7. **Step 5: Interpretation** The limit of $f(x)$ as $x$ approaches 5 is 75. **Final answer:** $$\lim_{x \to 5} \frac{x^3 - 125}{x - 5} = 75$$ This means as $x$ gets closer to 5, $f(x)$ approaches 75.