Question

If \[ \lim_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4} \] is finite, then \( a + b = \) __.

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

To solve trigonometric limits involving Taylor series, expand the functions and compare the powers of \( x \) to ensure the limit is finite.

Answer 1

The Correct Option is A

We are given the limit: \[ \lim_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4} \] To solve this, we first use the Taylor series expansions for \( \cos(2x) \) and \( \cos(4x) \) around \( x = 0 \): \[ \cos(2x) = 1 - 2x^2 + O(x^4) \] \[ \cos(4x) = 1 - 8x^2 + O(x^4) \] Substituting these expansions into the given expression: \[ \frac{\left( 1 - 2x^2 + O(x^4) \right) + a \left( 1 - 8x^2 + O(x^4) \right) - b}{x^4} \] Simplifying: \[ \frac{(1 + a - b) + (-2 + a(-8))x^2 + O(x^4)}{x^4} \] For this limit to be finite, the numerator must have no terms of degree less than \( x^4 \). Therefore, the coefficient of \( x^2 \) must be 0. Thus, we have: \[ -2 + a(-8) = 0 \quad \Rightarrow \quad a = \frac{2}{8} = \frac{1}{4} \] Additionally, for the constant term to cancel out, we must have: \[ 1 + a - b = 0 \quad \Rightarrow \quad 1 + \frac{1}{4} - b = 0 \quad \Rightarrow \quad b = \frac{5}{4} \] Therefore, \( a + b = \frac{1}{4} + \frac{5}{4} = 0 \). Thus, the correct answer is \( a + b = 0 \).