<p>\(\lim_{x \to \infty} \left(\frac{2^x+1}{2^x-1}\right)^{2^x} =\) ______</p>
Step-by-Step Solution
Key Concept: Rewrite the base as 1 + (a small term) to use the standard limit form (1 + 1/n)^n → e. Specifically, factor out the dominant term and express the exponent in terms of the reciprocal of what appears in the base adjustment.
<p><strong>Step 1:</strong> Recognize the indeterminate form (1)^∞. Rewrite the base:</p><p>$$\frac{2^x+1}{2^x-1} = \frac{2^x-1+2}{2^x-1} = 1 + \frac{2}{2^x-1}$$</p><p><strong>Step 2:</strong> Substitute into the original limit:</p><p>$$\lim_{x \to \infty} \left(1 + \frac{2}{2^x-1}\right)^{2^x}$$</p><p><strong>Step 3:</strong> Use the standard form $\lim_{n \to \infty} (1 + \frac{a}{n})^n = e^a$. Set $n = 2^x - 1$, so as $x \to \infty$, $n \to \infty$:</p><p>$$\lim_{x \to \infty} \left(1 + \frac{2}{2^x-1}\right)^{2^x} = \lim_{x \to \infty} \left(1 + \frac{2}{2^x-1}\right)^{2^x-1} \cdot \left(1 + \frac{2}{2^x-1}\right)^1$$</p><p><strong>Step 4:</strong> The first part approaches $e^2$ (by the standard limit with $a=2$), and the second part approaches 1:</p><p>$$e^2 \cdot 1 = e^2$$</p><p>∴ Answer: $e^2$</p>
Correct Answer: e^2