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Multiple Choice
Use series to evaluate the limit: .
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Verified step by step guidance
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Step 1: Recognize that the problem involves evaluating a limit as x approaches 0. To simplify the expressions, expand both the numerator and denominator using Taylor series approximations around x = 0.
Step 2: Expand \( \cos(2x) \) using its Taylor series: \( \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \dots \). Subtract \( \cos(2x) \) from 1 to get \( 1 - \cos(2x) = \frac{(2x)^2}{2!} - \frac{(2x)^4}{4!} + \dots \).
Step 3: Expand \( e^{2x} \) using its Taylor series: \( e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots \). Subtract \( e^{2x} \) from \( 1 + 2x \) to get \( 1 + 2x - e^{2x} = -\frac{(2x)^2}{2!} - \frac{(2x)^3}{3!} - \dots \).
Step 4: Substitute the Taylor series expansions for \( 1 - \cos(2x) \) and \( 1 + 2x - e^{2x} \) into the limit expression: \( \lim_{x \to 0} \frac{1 - \cos(2x)}{1 + 2x - e^{2x}} = \lim_{x \to 0} \frac{\frac{(2x)^2}{2!} - \frac{(2x)^4}{4!} + \dots}{-\frac{(2x)^2}{2!} - \frac{(2x)^3}{3!} - \dots} \).
Step 5: Simplify the fraction by canceling common terms in the numerator and denominator. Focus on the leading-order terms (the lowest power of x) in both the numerator and denominator, as higher-order terms become negligible as x approaches 0. This will allow you to evaluate the limit and determine the result.