Let's solve each problem step by step: --- ### **Problem 1** Compute the limit: \[ \lim_{x \to \infty} \frac{(\ln(x))^2}{\sqrt{x}} \] #### **Solution:** As \( x \to \infty \), both the numerator and denominator go to infinity. We can use **L'Hôpital's Rule** (since it's \(\infty/\infty\)): Let \( f(x) = (\ln x)^2 \), \( g(x) = \sqrt{x} \). Differentiate numerator and denominator: - \( f'(x) = 2\ln(x) \cdot \frac{1}{x} = \frac{2\ln(x)}{x} \) - \( g'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \) Apply L'Hôpital's Rule: \[ \lim_{x \to \infty} \frac{(\ln(x))^2}{\sqrt{x}} = \lim_{x \to \infty} \frac{2\ln(x)/x}{1/(2\sqrt{x})} = \lim_{x \to \infty} \frac{2\ln(x)}{x} \cdot 2\sqrt{x} = \lim_{x \to \infty} \frac{4\ln(x)\sqrt{x}}{x} = \lim_{x \to \infty} \frac{4\ln(x)}{\sqrt{x}} \] Still an \(\infty/\infty\) form, apply L'Hôpital's Rule again: - Numerator derivative: \( \frac{d}{dx}4\ln(x) = \frac{4}{x} \) - Denominator derivative: \( \frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}} \) \[ \lim_{x \to \infty} \frac{4\ln(x)}{\sqrt{x}} = \lim_{x \to \infty} \frac{4/x}{1/(2\sqrt{x})} = \lim_{x \to \infty} \frac{4}{x} \cdot 2\sqrt{x} = \lim_{x \to \infty} \frac{8\sqrt{x}}{x} = \lim_{x \to \infty} \frac{8}{\sqrt{x}} = 0 \] **Final Answer:** \[ \boxed{0} \] --- ### **Problem 2** Given \( y^2 + xy + x^2 = 1 \), find the equation of the tangent line at \( (0, -1) \) using implicit differentiation. #### **Solution:** Differentiate both sides with respect to \( x \): \[ 2y \frac{dy}{dx} + y + x \frac{dy}{dx} + 2x = 0 \] Group terms with \( \frac{dy}{dx} \): \[ 2y \frac{dy}{dx} + x \frac{dy}{dx} = -y - 2x \implies (2y + x)\frac{dy}{dx} = -y - 2x \implies \frac{dy}{dx} = \frac{-y - 2x}{2y + x} \] At \( (0, -1) \): \[ \frac{dy}{dx} = \frac{-(-1) - 2(0)}{2(-1) + 0} = \frac{1}{-2} = -\frac{1}{2} \] Equation of tangent line: \( y - y_0 = m(x - x_0) \) \[ y - (-1) = -\frac{1}{2}(x - 0) \implies y + 1 = -\frac{1}{2}x \implies y = -\frac{1}{2}x - 1 \] **Final Answer:** \[ \boxed{y = -\frac{1}{2}x - 1} \] --- ### **Problem 3** Biker \( A \) travels north at \( 6\,\text{m/s} \), Biker \( B \) east at \( 8\,\text{m/s} \). How fast is the distance between them changing one second before impact? #### **Solution:** Let \( x \) = eastward distance (B), \( y \) = northward distance (A), \( s \) = distance between them. \[ s^2 = x^2 + y^2 \] \[ \frac{ds}{dt} = \frac{1}{2s} \cdot 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s} \] Suppose they start at the intersection at \( t = 0 \), and impact at \( t = T \) (when both reach the intersection). One second before impact: \( t = T - 1 \). Let’s say at \( t \) seconds before impact, Biker A is \( y = 6t \) meters from the intersection and Biker B is \( x = 8t \) meters from the intersection. \[ s = \sqrt{(8t)^2 + (6t)^2} = \sqrt{64t^2 + 36t^2} = \sqrt{100t^2} = 10t \] \[ \frac{dx}{dt} = -8 \quad (\text{since x decreases as B approaches intersection}) \] \[ \frac{dy}{dt} = -6 \quad (\text{same for y}) \] \[ \frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s} \] At \( t = 1 \): \[ x = 8, \quad y = 6, \quad s = 10 \] \[ \frac{ds}{dt} = \frac{8(-8) + 6(-6)}{10} = \frac{-64 - 36}{10} = \frac{-100}{10} = -10\,\text{m/s} \] **Final Answer:** \[ \boxed{-10\,\text{m/s}} \] (The negative sign indicates the distance is decreasing.) --- ### **Problem 5** Find the absolute maximum and minimum of \( f(x) = x^3 + 3x^2 - 9x - 9 \) on \( -2 \leq x \leq 2 \). #### **Solution:** Find critical points: \[ f'(x) = 3x^2 + 6x - 9 \] Set \( f'(x) = 0 \): \[ 3x^2 + 6x - 9 = 0 \implies x^2 + 2x - 3 = 0 \implies (x + 3)(x - 1) = 0 \implies x = -3, \, 1 \] Only \( x = 1 \) is in \( [-2, 2] \). Check endpoints and critical point: - \( x = -2 \): \( f(-2) = (-8) + 12 + 18 - 9 = 13 \) - \( x = 1 \): \( f(1) = 1 + 3 - 9 - 9 = -14 \) - \( x = 2 \): \( f(2) = 8 + 12 - 18 - 9 = -7 \) Therefore: - **Absolute maximum:** \( \boxed{13} \) at \( x = -2 \) - **Absolute minimum:** \( \boxed{-14} \) at \( x = 1 \) --- #### **Summary of Final Answers:** 1. \( \boxed{0} \) 2. \( \boxed{y = -\frac{1}{2}x - 1} \) 3. \( \boxed{-10 \ \text{m/s}} \) 5. **Max:** \( \boxed{13} \) at \( x = -2 \), **Min:** \( \boxed{-14} \) at \( x = 1 \)