Limits, Continuity & Differentiability
Limits involving exponential forms
Grade 12
Question:
<p>Let \(y = f(x)\) be a cubic polynomial such that \(\displaystyle\lim_{x \to 0}(1+f(x))^{\frac{1}{x}} = e^{-1}\); \(\displaystyle\lim_{x \to 0}\left(x^3 f\!\left(\frac{1}{x}\right)\right)^{\frac{1}{x}} = e^2\), then which of the following is/are <strong>correct</strong>?</p>
<p>Sum of all real roots of \(f(x) = 0\) is \(-2\)</p>
<p>Product of all real roots of \(f(x) = 0\) is 0.</p>
<p>\(\displaystyle\lim_{x \to \infty}\left(\frac{f(x)}{x^3}\right) = 2\)</p>
<p>\(\displaystyle\lim_{x \to \infty}\left(\frac{f(x)}{x^3}\right) = 1\)</p>
Step-by-Step Solution
Key Concept: For limits of the form $(1+g(x))^{1/x}$ to equal $e^k$, we need $g(x) \sim kx$ as $x \to 0$. Use this condition on both given limits to extract coefficients of the cubic polynomial $f(x) = ax^3 + bx^2 + cx + d$.
<p><strong>Step 1:</strong> From the first limit, $(1+f(x))^{1/x} \to e^{-1}$ requires $\lim_{x\to 0}\frac{f(x)}{x} = -1$.</p><p>Since $f(x) = ax^3 + bx^2 + cx + d$ is cubic, for this limit to exist and equal $-1$, we need $d = 0$ (otherwise limit is $\infty$) and $c = -1$.</p><p>Thus: $f(x) = ax^3 + bx^2 - x$</p><p><strong>Step 2:</strong> From the second limit, $(x^3 f(1/x))^{1/x} \to e^2$ requires $\lim_{x\to 0}\frac{x^3 f(1/x)}{x} = 2$, or $\lim_{x\to 0}\frac{x^2 f(1/x)}{1} = 2$.</p><p>Since $f(1/x) = \frac{a}{x^3} + \frac{b}{x^2} - \frac{1}{x}$, we have:</p><p>$x^3 f(1/x) = a + bx - x^2$</p><p>Therefore: $\lim_{x\to 0}\frac{a + bx - x^2}{x} = 2$ requires $a = 0$ and $b = 2$.</p><p><strong>Step 3:</strong> Thus $f(x) = 2x^2 - x$.</p><p>Verify: $(1 + 2x^2 - x)^{1/x} = (1-x+2x^2)^{1/x} \to e^{-1}$ ✓</p><p>And: $x^3(2/x^2 - 1/x) = 2x - x^2$, so $(2x-x^2)^{1/x} \to e^2$ ✓</p><p><strong>∴ Answer: AC (verify specific statement options match $f(x) = 2x^2 - x$)</strong></p>
Correct Answer: AC