Certainly! I’ll answer **question 3a** step-by-step as requested. --- ## 3a. Find the limit: \[ \lim_{x \to } \frac{(2x^3) - x^2 - x^3}{x} \] --- ### **Step 1: Simplify the numerator** First, combine like terms in the numerator: \[ (2x^3) - x^2 - x^3 = (2x^3 - x^3) - x^2 = x^3 - x^2 \] So, the limit becomes: \[ \lim_{x \to } \frac{x^3 - x^2}{x} \] **Explanation:** We simplified the numerator by combining \(2x^3\) and \(-x^3\). --- ### **Step 2: Factor and reduce the expression** Factor \(x\) from the numerator: \[ x^3 - x^2 = x^2(x - 1) \] So, \[ \frac{x^3 - x^2}{x} = \frac{x^2(x - 1)}{x} = x(x - 1) \] Now take the limit as \(x \to \): \[ \lim_{x \to } x(x - 1) = \cdot ( - 1) = \] **Explanation:** Factoring allowed us to cancel \(x\) in the denominator, then direct substitution gives us the answer. --- ## **Final Answer** \[ \boxed{} \] Let me know if you want step-by-step solutions for another part!

# Complete Solutions to the Given Problems ## Problem 3a: Find the limit ### **Step 1: Simplify the expression** Given the limit: \[ \lim_{x \to 0} \frac{(2x^3) - x^2 - x^3}{x} \] First, combine like terms in the numerator: \[ (2x^3) - x^2 - x^3 = (2x^3 - x^3) - x^2 = x^3 - x^2 \] Thus, the limit can be rewritten as: \[ \lim_{x \to 0} \frac{x^3 - x^2}{x} \] **Explanation:** We simplified the numerator by combining \(2x^3\) and \(-x^3\). ### **Step 2: Factor and reduce the expression** Next, factor \(x\) from the numerator: \[ x^3 - x^2 = x^2(x - 1) \] Now the limit becomes: \[ \lim_{x \to 0} \frac{x^2(x - 1)}{x} = \lim_{x \to 0} x(x - 1) \] Now we can substitute \(x = 0\): \[ \lim_{x \to 0} x(x - 1) = 0 \cdot (0 - 1) = 0 \] **Explanation:** Factoring allowed us to cancel \(x\) in the denominator, and direct substitution gives us the answer. ## **Final Answer for Problem 3a** \[ \boxed{0} \] --- ## Problem 3b: Intermediate Value Theorem ### **Step 1: Establish the function and intervals** Let \(f(x) = x^3 - x^2 - 2\). We want to find if there is a root in the interval \([1, 2]\). Evaluate \(f(1)\) and \(f(2)\): - \(f(1) = 1^3 - 1^2 - 2 = 1 - 1 - 2 = -2\) - \(f(2) = 2^3 - 2^2 - 2 = 8 - 4 - 2 = 2\) ### **Step 2: Apply the Intermediate Value Theorem** Since \(f(1) < 0\) and \(f(2) > 0\), by the Intermediate Value Theorem, there exists at least one \(c\) in the interval \( (1, 2) \) such that \(f(c) = 0\). **Explanation:** The theorem states that if a continuous function takes on opposite signs at two points, it must cross zero in between. ## **Final Answer for Problem 3b** There is at least one root in the interval \([1, 2]\). --- ## Problem 4: Find the root using the theorem ### **Step 1: Define the function** Let \(f(x) = x^3 - x^2 - 2\). We already established that \(f(1) < 0\) and \(f(2) > 0\). ### **Step 2: Use numerical methods** Using numerical methods (like bisection), we can iteratively find the root: 1. Midpoint of \(1\) and \(2\): \(1.5\) - \(f(1.5) = (1.5)^3 - (1.5)^2 - 2 = 3.375 - 2.25 - 2 = -0.875\) (negative) 2. Midpoint of \(1.5\) and \(2\): \(1.75\) - \(f(1.75) = (1.75)^3 - (1.75)^2 - 2 = 5.359375 - 3.0625 - 2 = 0.296875\) (positive) 3. Midpoint of \(1.5\) and \(1.75\): \(1.625\) - \(f(1.625) = (1.625)^3 - (1.625)^2 - 2 \approx -0.397\) (negative) Continue this process to narrow down the root. **Explanation:** Using bisection, we iteratively find values that bracket the root until we reach a desired accuracy. ## **Final Answer for Problem 4** The root is approximately \(1.6\) (to one decimal place). --- ## Problem 5: Continuity and limits ### **Step 1: Define the function** Let \(g(x) = |x| + [x]\), where \([x]\) is the greatest integer function. ### **Step 2: Analyze continuity** - On intervals \( (0, 2) \), check points \(x = 0\) and \(x = 2\). At \(x = 0\): - Left limit \(g(0^-) = 0 + 0 = 0\) - Right limit \(g(0^+) = 0 + 0 = 0\) - \(g(0) = 0\) (continuous) At \(x = 2\): - Left limit \(g(2^-) = 2 + 1 = 3\) - Right limit \(g(2^+) = 2 + 2 = 4\) \(g(2)\) is not equal to the limits, so it’s discontinuous at \(x = 2\). **Explanation:** A function is continuous at a point if the left and right limits equal the function value at that point. ## **Final Answer for Problem 5** - \(g(x)\) is continuous at \(x = 0\) and discontinuous at \(x = 2\). --- Feel free to ask for more detailed explanations or additional problems!