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

The provided limit is: \[ \lim_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4} \] Using Taylor series expansions around \( x = 0 \): \[ \cos(2x) = 1 - 2x^2 + O(x^4) \] \[ \cos(4x) = 1 - 8x^2 + O(x^4) \] Substituting these into the limit expression: \[ \frac{\left( 1 - 2x^2 + O(x^4) \right) + a \left( 1 - 8x^2 + O(x^4) \right) - b}{x^4} \] Simplifying the numerator: \[ \frac{(1 + a - b) + (-2 - 8a)x^2 + O(x^4)}{x^4} \] For the limit to be finite, terms with powers of \( x \) less than 4 in the numerator must be zero. This requires the coefficient of \( x^2 \) to be 0: \[ -2 - 8a = 0 \quad \Rightarrow \quad a = -\frac{2}{8} = -\frac{1}{4} \] Additionally, the constant term must also be zero: \[ 1 + a - b = 0 \quad \Rightarrow \quad 1 - \frac{1}{4} - b = 0 \quad \Rightarrow \quad b = \frac{3}{4} \] Therefore, \( a + b = -\frac{1}{4} + \frac{3}{4} = \frac{2}{4} = \frac{1}{2} \). Thus, the correct answer is \( a + b = \frac{1}{2} \).