1. **Problem statement:** Evaluate the following limits and analyze the given functions and expressions step-by-step. --- 2. **Limit 1:** $$\lim_{x \to 1} \frac{\sqrt[3]{x} + 7 - 2}{x - 1}$$ - Substitute $x=1$: numerator $= \sqrt[3]{1} + 7 - 2 = 1 + 7 - 2 = 6$ - Denominator $= 1 - 1 = 0$ - This is a form $\frac{6}{0}$ which tends to infinity or does not exist. - Check left and right limits: - As $x \to 1^-$, denominator $<0$, fraction tends to $-\infty$ - As $x \to 1^+$, denominator $>0$, fraction tends to $+\infty$ - **Conclusion:** Limit does not exist. --- 3. **Limit 2:** $$\lim_{x \to +\infty} \frac{x - \sqrt[3]{x^2}}{x}$$ - Rewrite numerator: $x - x^{2/3}$ - Divide numerator and denominator by $x$: $$\frac{x - x^{2/3}}{x} = 1 - x^{-1/3}$$ - As $x \to +\infty$, $x^{-1/3} \to 0$ - So limit $= 1 - 0 = 1$ --- 4. **Limit 3:** $$\lim_{x \to -\infty} \sqrt[3]{x^3} - x + 3x$$ - Note $\sqrt[3]{x^3} = x$ - Expression simplifies to $x - x + 3x = 3x$ - As $x \to -\infty$, $3x \to -\infty$ - **Limit:** $-\infty$ --- 5. **Limit 4:** Same as Limit 2, so answer is $1$. --- 6. **Expression A:** $$A = \frac{\sqrt[4]{9} \cdot \sqrt[3]{3} \cdot \sqrt[3]{9}}{\sqrt[5]{81} \cdot x \cdot \sqrt{\sqrt{3}}}$$ - Simplify numerator: - $\sqrt[4]{9} = 9^{1/4} = (3^2)^{1/4} = 3^{1/2} = \sqrt{3}$ - $\sqrt[3]{3} = 3^{1/3}$ - $\sqrt[3]{9} = 9^{1/3} = (3^2)^{1/3} = 3^{2/3}$ - Numerator combined: $\sqrt{3} \cdot 3^{1/3} \cdot 3^{2/3} = 3^{1/2} \cdot 3^{1/3 + 2/3} = 3^{1/2 + 1} = 3^{3/2}$ - Denominator: - $\sqrt[5]{81} = 81^{1/5} = (3^4)^{1/5} = 3^{4/5}$ - $\sqrt{\sqrt{3}} = (3^{1/2})^{1/2} = 3^{1/4}$ - Denominator combined: $3^{4/5} \cdot x \cdot 3^{1/4} = x \cdot 3^{4/5 + 1/4} = x \cdot 3^{(16/20 + 5/20)} = x \cdot 3^{21/20}$ - So, $$A = \frac{3^{3/2}}{x \cdot 3^{21/20}} = \frac{3^{30/20}}{x \cdot 3^{21/20}} = \frac{3^{9/20}}{x}$$ --- 7. **Expression B:** $$B = \frac{\sqrt[15]{35} \cdot \sqrt[3]{9} \cdot \sqrt[5]{93}}{\sqrt[5]{3}}$$ - Simplify each term: - $\sqrt[15]{35} = 35^{1/15}$ - $\sqrt[3]{9} = 9^{1/3} = (3^2)^{1/3} = 3^{2/3}$ - $\sqrt[5]{93} = 93^{1/5}$ - $\sqrt[5]{3} = 3^{1/5}$ - Combine numerator powers of 3: - $3^{2/3} \cdot 93^{1/5} \cdot 35^{1/15}$ (cannot simplify 35 and 93 further) - Divide by $3^{1/5}$: $$B = 35^{1/15} \cdot 93^{1/5} \cdot 3^{2/3 - 1/5} = 35^{1/15} \cdot 93^{1/5} \cdot 3^{(10/15 - 3/15)} = 35^{1/15} \cdot 93^{1/5} \cdot 3^{7/15}$$ --- 8. **Function:** $$f(x) = \frac{x^2}{x - 1}$$ - Domain: $x \neq 1$ - Derivative given: $$f'(x) = \frac{x^2 - 2x}{(x - 1)^2}$$ --- 9. **Inverse function values:** - Given $g^{-1}(4 \sqrt[3]{5})$, $g^{-1}(4 \sqrt{2})$, $g^{-1}(4 \sqrt[6]{10})$ - Without explicit $g$, cannot find exact values. --- 10. **Equation (E):** $$x^3 = 3x - 3$$ - Has one unique real root $\alpha$. --- 11. **Function at root:** - $$f(\alpha) = \frac{3}{\alpha}$$ - Given inequality: $$-\frac{3}{2} < f(\alpha) < -1$$ --- 12. **Function:** $$f(x) = \sqrt[3]{x} + 1$$ - Inverse function $f^{-1}(y)$ satisfies: $$y = \sqrt[3]{x} + 1 \implies \sqrt[3]{x} = y - 1 \implies x = (y - 1)^3$$ - So, $$f^{-1}(y) = (y - 1)^3$$ --- **Final answers:** - Limit 1: Does not exist - Limit 2: $1$ - Limit 3: $-\infty$ - Limit 4: $1$ - Expression A: $\frac{3^{9/20}}{x}$ - Expression B: $35^{1/15} \cdot 93^{1/5} \cdot 3^{7/15}$ - $f'(x) = \frac{x^2 - 2x}{(x - 1)^2}$ - $f^{-1}(y) = (y - 1)^3$