Let's solve the problems step by step as requested: --- ## **1. (20 points)** ### (a) Find the following limits: #### (i) \(\lim_{x \to 0} \frac{\sin 3x}{\sin 5x}\) **Step 1:** Recall that \(\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1\). \[ \frac{\sin 3x}{\sin 5x} = \frac{\sin 3x}{3x} \cdot \frac{5x}{\sin 5x} \cdot \frac{3x}{5x} \] **Step 2:** Take the limit: \[ \lim_{x \to 0} \frac{\sin 3x}{\sin 5x} = \lim_{x \to 0} \left(\frac{\sin 3x}{3x} \cdot \frac{5x}{\sin 5x} \cdot \frac{3}{5}\right) \] Each limit: - \(\lim_{x \to 0} \frac{\sin 3x}{3x} = 1\) - \(\lim_{x \to 0} \frac{5x}{\sin 5x} = 1\) So, \[ \lim_{x \to 0} \frac{\sin 3x}{\sin 5x} = 1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5} \] **Final answer:** \[ \boxed{\frac{3}{5}} \] --- #### (ii) \(\lim_{x \to 2} \frac{x^2 - 1}{x^2(x^2-4)}\) **Step 1:** Substitute \(x = 2\): - \(x^2 - 1 = 4 - 1 = 3\) - \(x^2 = 4\) - \(x^2 - 4 = 0\) So denominator is zero: \(4 \times 0 = 0\), numerator is 3. So, the function approaches infinity or negative infinity. Let's check the sign: - For \(x \to 2^+\): \(x^2-4\) is slightly positive, denominator positive, numerator positive: \(+\infty\) - For \(x \to 2^-\): \(x^2-4\) is slightly negative, denominator negative, numerator positive: \(-\infty\) **So the limit does not exist (infinite discontinuity).** **Final Answer:** \[ \boxed{\text{The limit does not exist (approaches } +\infty \text{ from right and } -\infty \text{ from left)}} \] --- ### (b) State (but do not prove) the epsilon-delta definition of \(\lim_{x \to 0} \sin(5x) = 0\): **Answer:** For every \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x\), \[ 0 < |x| < \delta \implies |\sin(5x) - 0| < \epsilon \] --- ## **2. (20 points)** Use the Intermediate Value Theorem to show that \(e^x = 4 - x^2\) has a solution in \([0,2]\). **Step 1:** Define \(f(x) = e^x - 4 + x^2\). **Step 2:** Check values at the endpoints: - \(f(0) = 1 - 4 + 0 = -3\) - \(f(2) = e^2 - 4 + 4 = e^2 \approx 7.389 > 0\) **Step 3:** Since \(f(0) < 0\) and \(f(2) > 0\), by the Intermediate Value Theorem, \(f(x)\) has a root in \((0,2)\). **Final Answer:** By the intermediate value theorem, there exists \(c \in (0,2)\) such that \(e^c = 4 - c^2\). --- ## **3. (20 points)** ### (a) Is there any relationship between the slope of the tangent line to a curve \(f\) at a point \(a\) and the derivative \(f'(a)\)? **Answer:** Yes. The slope of the tangent line to the curve \(y=f(x)\) at the point \(x=a\) is exactly the value of the derivative at that point, \(f'(a)\). --- ### (b) Consider \(f(x) = 3x^2 - 2x + 1\). Find the equation of the tangent line at \(x = -1\). **Step 1:** Calculate \(f(-1)\): \[ f(-1) = 3(-1)^2 - 2(-1) + 1 = 3(1) + 2 + 1 = 6 \] **Step 2:** Find \(f'(x)\): \[ f'(x) = 6x - 2 \] \[ f'(-1) = 6(-1) - 2 = -6 - 2 = -8 \] **Step 3:** Equation of tangent line: \[ y - f(-1) = f'(-1)(x + 1) \] \[ y - 6 = -8(x + 1) \implies y = -8x - 8 + 6 \implies y = -8x - 2 \] **Final answer:** \[ \boxed{y = -8x - 2} \] --- ## **4. (30 points)** Suppose \(x, y\) are differentiable functions of \(t\), and \(x^2 + y^2 - 3y = 0\). Use implicit differentiation to find \(\frac{dy}{dt}\). (No need to simplify.) **Differentiate both sides with respect to \(t\):** \[ \frac{d}{dt}(x^2 + y^2 - 3y) = 0 \] \[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} - 3\frac{dy}{dt} = 0 \] \[ 2x\frac{dx}{dt} + (2y - 3)\frac{dy}{dt} = 0 \] **Solve for \(\frac{dy}{dt}\):** \[ (2y - 3)\frac{dy}{dt} = -2x\frac{dx}{dt} \] \[ \frac{dy}{dt} = \frac{-2x}{2y - 3}\frac{dx}{dt} \] **Final answer:** \[ \boxed{\frac{dy}{dt} = \frac{-2x}{2y - 3} \frac{dx}{dt}} \] --- ## **5. (10 points)** Using the Bisection method, approximate \(2^{1/3}\) with error less than \(\frac{1}{2}\). Let \(f(x) = x^3 - 2\). - \(f(1) = 1 - 2 = -1\) - \(f(2) = 8 - 2 = 6\) So root is in \([1,2]\). **Bisection steps:** 1. \(x_1 = 1\), \(x_2 = 2\) 2. \(x_m = 1.5\), \(f(1.5) = (1.5)^3 - 2 = 3.375 - 2 = 1.375\) (root is between 1 and 1.5) 3. \(x_m = 1.25\), \(f(1.25) = (1.25)^3 - 2 = 1.953125 - 2 = -0.046875\) (root is between 1.25 and 1.5) 4. \(x_m = 1.375\), \(f(1.375) = (1.375)^3 - 2 = 2.595703125 - 2 = 0.595703125\) (root is between 1.25 and 1.375) 5. \(x_m = 1.3125\), \(f(1.3125) = (1.3125)^3 - 2 = 2.260986328125 - 2 = 0.260986328125\) (root is between 1.25 and 1.3125) The interval is \([1.25, 1.3125]\), with width \(0.0625\), less than \(0.5\). **Approximate value:** The midpoint \(x \approx 1.3125\). **Final answer:** \[ \boxed{2^{1/3} \approx 1.3125 \text{ (with error less than } \frac{1}{2})} \] --- Let me know if you need further clarifications on any part!

Certainly! Here are the final answers for each part: 1. **Limit \(\lim_{x \to 0} \frac{\sin 3x}{\sin 5x}\):** \(\boxed{\frac{3}{5}}\) 2. **Limit \(\lim_{x \to 2} \frac{x^2 - 1}{x^2(x^2-4)}\):** Does not exist (approaches \(\pm \infty\)). 3. **Epsilon-delta definition of \(\lim_{x \to 0} \sin(5x) = 0\):** For every \(\epsilon > 0\), there exists \(\delta > 0\) such that if \(0 < |x| < \delta\), then \(|\sin(5x)| < \epsilon\). 4. **Existence of solution to \(e^x = 4 - x^2\) in \([0,2]\):** Yes, there is at least one solution in \((0,2)\). 5. **Relationship between slope and derivative:** Yes, the slope of the tangent line at a point is \(f'(a)\). 6. **Tangent line to \(f(x) = 3x^2 - 2x + 1\) at \(x=-1\):** \[ \boxed{y = -8x - 2} \] 7. **Implicit differentiation result for \(x^2 + y^2 - 3y= 0\):** \[ \boxed{\frac{dy}{dt} = \frac{-2x}{2y - 3} \frac{dx}{dt}} \] 8. **Approximation of \(2^{1/3}\) via Bisection with error < 0.5:** \(\boxed{1.3125}\) Let me know if you'd like any further details!