Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>The value of \(\displaystyle\lim_{x \to \infty} \frac{e^x\left[\left(2^{x^n}\right)^{1/e^x} - \left(e^{x^n}\right)^{1/e^x}\right]}{x^n}\) where \(n\) is a positive integer, is:</p>
<p>\(\ln 2 - \ln 3\)</p>
<p>\(\ln 3 - \ln 2\)</p>
<p>\(0\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Rewrite the exponential terms using $a^{1/e^x} = e^{\ln a/e^x}$ and apply Taylor expansion $e^u \approx 1 + u$ for small $u$ as $x \to \infty$, where $u = \ln a/e^x \to 0$.
<p><strong>Step 1:</strong> Rewrite using logarithmic form: $(2^{x^n})^{1/e^x} = e^{\frac{x^n \ln 2}{e^x}}$ and $(e^{x^n})^{1/e^x} = e^{\frac{x^n}{e^x}}$</p><p><strong>Step 2:</strong> As $x \to \infty$, both exponents $\frac{x^n \ln 2}{e^x}$ and $\frac{x^n}{e^x} \to 0$ (exponential dominates polynomial). Use Taylor expansion: $e^u \approx 1 + u$ for small $u$.</p><p><strong>Step 3:</strong> Apply expansion:</p><p>$(2^{x^n})^{1/e^x} \approx 1 + \frac{x^n \ln 2}{e^x}$</p><p>$(e^{x^n})^{1/e^x} \approx 1 + \frac{x^n}{e^x}$</p><p><strong>Step 4:</strong> Compute the difference:</p><p>$(2^{x^n})^{1/e^x} - (e^{x^n})^{1/e^x} \approx \frac{x^n(\ln 2 - 1)}{e^x}$</p><p><strong>Step 5:</strong> Substitute into the limit:</p><p>$$\lim_{x \to \infty} \frac{e^x \cdot \frac{x^n(\ln 2 - 1)}{e^x}}{x^n} = \lim_{x \to \infty} \frac{x^n(\ln 2 - 1)}{x^n} = \ln 2 - 1$$</p><p>∴ <strong>Answer: B</strong> (which equals $\ln 2 - 1$ or $\ln(2/e)$)</p>
Correct Answer: B