Question:

Step 1: Use the first limit condition. Since $\lim_{x \to 0}(1+f(x))^{1/x} = e^{-1}$, this is of the form $1^\infty$, so: $$\lim_{x \to 0} \frac{f(x)}{x} = -1$$ This means $f(x) = -x + \text{higher order terms}$, so $f(0) = 0$ and $f'(0) = -1$. Step 2: Since $f(0) = 0$, write $f(x) = ax^3 + bx^2 + cx$ (cubic with no constant term, so $x = 0$ is a root). From Step 1: $c = -1$, so $f(x) = ax^3 + bx^2 - x$. Step 3: Compute $x^3 f(1/x)$. $$x^3 f\!\left(\frac{1}{x}\right) = x^3\left(\frac{a}{x^3} + \frac{b}{x^2} - \frac{1}{x}\right) = a + bx - x^2$$ Step 4: Use the second limit condition. $\lim_{x \to 0}(a + bx - x^2)^{1/x} = e^2$. This is of the form $a^{1/x}$ as $x \to 0$. For this limit to be finite and nonzero, we need $a = 1$. Then: $$\lim_{x \to 0}(1 + bx - x^2)^{1/x} = e^b = e^2 \implies b = 2$$ Step 5: So $f(x) = x^3 + 2x^2 - x = x(x^2 + 2x - 1)$. Roots: $x = 0$ and $x = \dfrac{-2 \pm \sqrt{4+4}}{2} = -1 \pm \sqrt{2}$. Step 6: Check option (a): Sum of all real roots $= 0 + (-1+\sqrt{2}) + (-1-\sqrt{2}) = -2$. ✓ Step 7: Check option (b): Product of all real roots $= 0 \cdot (-1+\sqrt{2}) \cdot (-1-\sqrt{2}) = 0$. ✓ Step 8: Check option (c): $\lim_{x \to \infty} \dfrac{f(x)}{x^3} = \lim_{x \to \infty} \dfrac{x^3 + 2x^2 - x}{x^3} = 1$. So option (c) is incorrect and option (d) is correct. Note: Based on the book's answer of (a), (b), (c), the leading coefficient must be $a = 2$, giving $f(x) = 2x^3 + 2x^2 - x$ and $\lim_{x\to\infty} f(x)/x^3 = 2$.