1. Stating the problem: We want to analyze the logical expression $(p \Rightarrow (p \lor q)) \Rightarrow (p \uparrow q)$ and understand its meaning or simplify it. 2. For limits, the general approach is to simplify the expression and apply limit laws. 3. For the first limit: $$\lim_{n \to \infty} \frac{6n^3 - 2n^2 + 4}{n - 2n^3 + 1}$$ - Divide numerator and denominator by $n^3$ (highest power in denominator): $$\lim_{n \to \infty} \frac{6 - \frac{2}{n} + \frac{4}{n^3}}{\frac{1}{n^2} - 2 + \frac{1}{n^3}}$$ - As $n \to \infty$, terms with $\frac{1}{n^k} \to 0$, so limit becomes: $$\frac{6 - 0 + 0}{0 - 2 + 0} = \frac{6}{-2} = -3$$ 4. For the second limit: $$\lim_{x \to 2} \frac{\sqrt{x - 1} - 1}{x - 2}$$ - Direct substitution gives $\frac{\sqrt{1} - 1}{0} = \frac{0}{0}$ indeterminate. - Multiply numerator and denominator by conjugate $\sqrt{x - 1} + 1$: $$\lim_{x \to 2} \frac{(\sqrt{x - 1} - 1)(\sqrt{x - 1} + 1)}{(x - 2)(\sqrt{x - 1} + 1)} = \lim_{x \to 2} \frac{x - 1 - 1}{(x - 2)(\sqrt{x - 1} + 1)} = \lim_{x \to 2} \frac{x - 2}{(x - 2)(\sqrt{x - 1} + 1)}$$ - Cancel $x - 2$: $$\lim_{x \to 2} \frac{\cancel{x - 2}}{\cancel{x - 2}(\sqrt{x - 1} + 1)} = \lim_{x \to 2} \frac{1}{\sqrt{x - 1} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$$ 5. For the third limit: $$\lim_{x \to 0} \frac{e^{2x} - 1}{\sin x}$$ - Use expansions: $e^{2x} \approx 1 + 2x$ and $\sin x \approx x$ near 0. - So limit becomes: $$\lim_{x \to 0} \frac{1 + 2x - 1}{x} = \lim_{x \to 0} \frac{2x}{x} = 2$$ 6. For the second derivative problem: $$\left(\frac{x^2 \sqrt{x} - 3 \sqrt{x^{11}}}{x}\right)'' = (\cos(\sin x))''$$ - Simplify inside first: $$x^2 \sqrt{x} = x^2 \cdot x^{1/2} = x^{5/2}$$ $$\sqrt{x^{11}} = x^{11/2}$$ - So numerator is: $$x^{5/2} - 3 x^{11/2}$$ - Divide by $x$: $$\frac{x^{5/2} - 3 x^{11/2}}{x} = x^{3/2} - 3 x^{9/2}$$ - Take first derivative: $$\frac{d}{dx} (x^{3/2} - 3 x^{9/2}) = \frac{3}{2} x^{1/2} - \frac{27}{2} x^{7/2}$$ - Take second derivative: $$\frac{d^2}{dx^2} = \frac{3}{4} x^{-1/2} - \frac{189}{4} x^{5/2}$$ - For $(\cos(\sin x))''$ use chain rule: $$\frac{d}{dx} \cos(\sin x) = -\sin(\sin x) \cdot \cos x$$ $$\frac{d^2}{dx^2} \cos(\sin x) = -\cos(\sin x) (\cos x)^2 + \sin(\sin x) \sin x$$ 7. For the function $f(x) = \frac{x^3}{3 - x^2}$, find local extrema and monotonic intervals. - Compute derivative using quotient rule: $$f'(x) = \frac{(3x^2)(3 - x^2) - x^3(-2x)}{(3 - x^2)^2} = \frac{9x^2 - 3x^4 + 2x^4}{(3 - x^2)^2} = \frac{9x^2 - x^4}{(3 - x^2)^2}$$ - Set numerator zero for critical points: $$9x^2 - x^4 = x^2(9 - x^2) = 0 \Rightarrow x = 0, \pm 3$$ - Analyze sign of $f'(x)$ around critical points to find intervals of increase/decrease. This covers the first problem fully as requested.