1. **State the problem:** We want to find the limits of various functions as $x \to \infty$ or $x \to 0$ based on the expressions and descriptions given. 2. **Analyze the first limit:** $$\lim_{x \to \infty} \frac{x + 2 \sin(3x) + 2x}{x + x^2 - 1 + x^9 + 2x^2}$$ 3. **Simplify numerator and denominator:** Numerator: $x + 2 \sin(3x) + 2x = 3x + 2 \sin(3x)$ Denominator: $x + x^2 - 1 + x^9 + 2x^2 = x^9 + 3x^2 + x - 1$ 4. **Identify dominant terms as $x \to \infty$:** In numerator, dominant term is $3x$. In denominator, dominant term is $x^9$. 5. **Divide numerator and denominator by $x^9$ to find limit:** $$\lim_{x \to \infty} \frac{3x + 2 \sin(3x)}{x^9 + 3x^2 + x - 1} = \lim_{x \to \infty} \frac{\frac{3x}{x^9} + \frac{2 \sin(3x)}{x^9}}{\frac{x^9}{x^9} + \frac{3x^2}{x^9} + \frac{x}{x^9} - \frac{1}{x^9}} = \lim_{x \to \infty} \frac{\frac{3}{x^8} + \frac{2 \sin(3x)}{x^9}}{1 + \frac{3}{x^7} + \frac{1}{x^8} - \frac{1}{x^9}}$$ 6. **Evaluate the limit:** As $x \to \infty$, terms with $x$ in denominator go to zero, so numerator $\to 0$, denominator $\to 1$. Therefore, $$\lim_{x \to \infty} \frac{x + 2 \sin(3x) + 2x}{x + x^2 - 1 + x^9 + 2x^2} = 0$$ 7. **Analyze the second limit:** $$\lim_{x \to \infty} \frac{x + x^2 - 1 + x^9 + 2x^2}{\sqrt{x + x^2}}$$ 8. **Simplify numerator and denominator:** Numerator: $x^9 + x^2 + 2x^2 + x - 1 = x^9 + 3x^2 + x - 1$ Denominator: $\sqrt{x + x^2} = \sqrt{x^2 + x} = x \sqrt{1 + \frac{1}{x}}$ 9. **Divide numerator and denominator by $x^9$ and $x$ respectively:** $$\lim_{x \to \infty} \frac{x^9 + 3x^2 + x - 1}{x \sqrt{1 + \frac{1}{x}}} = \lim_{x \to \infty} \frac{x^9}{x} \cdot \frac{1 + \frac{3}{x^7} + \frac{1}{x^8} - \frac{1}{x^9}}{\sqrt{1 + \frac{1}{x}}} = \lim_{x \to \infty} x^8 \cdot \frac{1 + 0 + 0 - 0}{1} = \infty$$ 10. **Analyze the third limit:** $$\lim_{x \to \infty} \frac{x \sin(1)}{x}$$ 11. **Simplify:** $$\frac{x \sin(1)}{x} = \sin(1)$$ 12. **Evaluate:** Since $\sin(1)$ is a constant, $$\lim_{x \to \infty} \frac{x \sin(1)}{x} = \sin(1) \approx 0.8415$$ 13. **Analyze the limit of $x \sin(1/x)$ as $x \to \infty$ and $x \to 0$:** - As $x \to \infty$, $1/x \to 0$, so $\sin(1/x) \approx 1/x$. - Therefore, $x \sin(1/x) \approx x \cdot \frac{1}{x} = 1$. - As $x \to 0$, $1/x \to \infty$, and $\sin(1/x)$ oscillates between $-1$ and $1$, so $x \sin(1/x) \to 0$ because $x \to 0$. **Final answers:** - $$\lim_{x \to \infty} \frac{x + 2 \sin(3x) + 2x}{x + x^2 - 1 + x^9 + 2x^2} = 0$$ - $$\lim_{x \to \infty} \frac{x + x^2 - 1 + x^9 + 2x^2}{\sqrt{x + x^2}} = \infty$$ - $$\lim_{x \to \infty} \frac{x \sin(1)}{x} = \sin(1)$$ - $$\lim_{x \to \infty} x \sin(1/x) = 1$$ - $$\lim_{x \to 0} x \sin(1/x) = 0$$