<p>\(\lim_{x \to 0} \left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{x^2}} =\) ______</p>
Step-by-Step Solution
Key Concept: Recognize this as an indeterminate form 1^∞ by rewriting the base as 1 + (difference term), then use the standard limit formula: lim(1 + f(x))^(1/f(x)) = e when f(x) → 0.
<p><strong>Step 1:</strong> Identify the form. As x → 0: base → 1 and exponent → ∞, giving 1^∞ (indeterminate).</p><p><strong>Step 2:</strong> Rewrite the base as 1 plus a correction term:</p><p>$$\frac{1+5x^2}{1+3x^2} = \frac{1+3x^2 + 2x^2}{1+3x^2} = 1 + \frac{2x^2}{1+3x^2}$$</p><p><strong>Step 3:</strong> Take logarithm of the limit:</p><p>$$\ln L = \lim_{x \to 0} \frac{1}{x^2} \ln\left(1 + \frac{2x^2}{1+3x^2}\right)$$</p><p><strong>Step 4:</strong> Use ln(1 + u) ≈ u for small u:</p><p>$$\ln L = \lim_{x \to 0} \frac{1}{x^2} \cdot \frac{2x^2}{1+3x^2} = \lim_{x \to 0} \frac{2}{1+3x^2} = 2$$</p><p><strong>Step 5:</strong> Therefore: L = e^2</p><p>∴ Answer: <strong>e²</strong></p>
Correct Answer: e^2