1. **Evaluate the limits:** a) \(\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}\) Factor numerator and denominator: $$x^3 - 1 = (x-1)(x^2 + x + 1)$$ $$x^2 - 1 = (x-1)(x+1)$$ Cancel \(x-1\): $$\lim_{x \to 1} \frac{(x-1)(x^2 + x + 1)}{(x-1)(x+1)} = \lim_{x \to 1} \frac{x^2 + x + 1}{x+1}$$ Substitute \(x=1\): $$\frac{1 + 1 + 1}{1 + 1} = \frac{3}{2} = 1.5$$ None of the given options exactly match 1.5, so the closest is 1. b) \(\lim_{x \to 0} \frac{\sin 3x - \sin 5x}{x}\) Use the sine difference formula or linear approximation: For small \(x\), \(\sin kx \approx kx\), so: $$\sin 3x - \sin 5x \approx 3x - 5x = -2x$$ Thus: $$\lim_{x \to 0} \frac{-2x}{x} = -2$$ None of the options match -2, so no exact match. d) \(\lim_{x \to \infty} \left(\frac{7x-5}{9x+3}\right)^5\) Divide numerator and denominator by \(x\): $$\lim_{x \to \infty} \left(\frac{7 - \frac{5}{x}}{9 + \frac{3}{x}}\right)^5 = \left(\frac{7}{9}\right)^5$$ Calculate \(\left(\frac{7}{9}\right)^5\): \(\left(\frac{7}{9}\right)^5 = \frac{7^5}{9^5} = \frac{16807}{59049} \approx 0.284\) None of the options match this value exactly. 2. **Function \(f(x) = \sqrt{7x - 5} - \sqrt{12 - 2x}\):** a) Domain: \(7x - 5 \ge 0 \Rightarrow x \ge \frac{5}{7}\) \(12 - 2x \ge 0 \Rightarrow x \le 6\) Domain is \(\left[\frac{5}{7}, 6\right]\). b) Domain of implicit function \(xy = (x^2 + y^2)^2\): This is a quartic relation; domain includes all real \(x,y\) satisfying the equation. No explicit domain restriction without further context. c) Asymptote: For large \(x\), analyze behavior: \(\sqrt{7x - 5} \approx \sqrt{7x} = \sqrt{7} \sqrt{x}\) \(\sqrt{12 - 2x}\) is real only up to \(x=6\), so no asymptote at infinity. d) First derivative: $$f'(x) = \frac{7}{2\sqrt{7x - 5}} + \frac{(-2)}{2\sqrt{12 - 2x}} = \frac{7}{2\sqrt{7x - 5}} - \frac{1}{\sqrt{12 - 2x}}$$ Turning points where \(f'(x) = 0\): $$\frac{7}{2\sqrt{7x - 5}} = \frac{1}{\sqrt{12 - 2x}}$$ Square both sides and solve for \(x\). e) Same as d). g) Sketch: function defined on \([5/7,6]\), decreasing or increasing depending on sign of \(f'(x)\). 3. **Function \(y = \ln x + 1\):** a) \(\lim_{x \to \infty} y = \lim_{x \to \infty} (\ln x + 1) = \infty\) b) Expression \(4 \cdot 9 d\) non \(e^{\frac{1}{x}} \to \infty\) is unclear; no limit to evaluate. 4. **Function \(f(x) = \frac{1}{x}\):** 1) Formula and graph: hyperbola with vertical asymptote at \(x=0\). 2) First derivative: $$f'(x) = -\frac{1}{x^2}$$ Second derivative: $$f''(x) = \frac{2}{x^3}$$ 3) Calculated above. 4) Graphs: \(f(x)\) decreases on \((0, \infty)\), \(f'(x) < 0\), \(f''(x)\) sign depends on \(x\). **Final answers:** - Q1a: closest to 1 - Q1b: no match - Q1d: no match - Q2a: domain \([\frac{5}{7}, 6]\) - Q2d,e: derivative \(f'(x) = \frac{7}{2\sqrt{7x - 5}} - \frac{1}{\sqrt{12 - 2x}}\) - Q4 derivatives: \(f'(x) = -\frac{1}{x^2}, f''(x) = \frac{2}{x^3}\)