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 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 expressions 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 exponential form:</p><p>$(2^{x^n})^{1/e^x} = e^{x^n \ln 2/e^x}$ and $(e^{x^n})^{1/e^x} = e^{x^n/e^x}$</p><p><strong>Step 2:</strong> Factor and simplify:</p><p>$$e^x\left[e^{x^n\ln 2/e^x} - e^{x^n/e^x}\right] = e^x \cdot e^{x^n/e^x}\left[e^{x^n(\ln 2 - 1)/e^x} - 1\right]$$</p><p><strong>Step 3:</strong> Apply Taylor expansion $e^u - 1 \approx u$ for small $u$:</p><p>$$e^{x^n/e^x}\left[e^{x^n(\ln 2 - 1)/e^x} - 1\right] \approx e^{x^n/e^x} \cdot \frac{x^n(\ln 2 - 1)}{e^x}$$</p><p><strong>Step 4:</strong> Since $e^{x^n/e^x} \to 1$ as $x \to \infty$ (as $x^n/e^x \to 0$):</p><p>$$\lim_{x \to \infty} \frac{e^x \cdot \frac{x^n(\ln 2 - 1)}{e^x}}{x^n} = \ln 2 - 1$$</p><p>∴ Answer: <strong>B</strong> (which is $\ln 2 - 1$)</p>
Correct Answer: B