Question

If $ \lim_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4} $ is finite, then $ (a + b) $ is equal to:

• A. \( \frac{1}{2} \)
• B. 0
• C. \( \frac{3}{4} \)
• D. -1

Hint:

In limits involving higher powers, use Taylor expansions to approximate the terms. This helps to identify the coefficients that must satisfy the condition for the limit to exist.

Answer 1

The Correct Option is A

We begin by expanding the terms in the given expression. Using the Taylor series expansion of \( \cos x \) around \( x = 0 \), we have: \[ \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + O(x^6) \] \[ \cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} + O(x^6) \] Thus, the given expression becomes: \[ L = \lim_{x \to 0} \frac{ \left( 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} \right) + a \left( 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} \right) - b }{x^4} \] Simplifying the expression: \[ L = \frac{1 + a - b}{x^4} + \frac{a \cdot \frac{(4x)^2}{2!} + O(x^4)}{x^4} + \text{higher order terms} \] For the limit to be finite, the coefficient of \( x^4 \) in the numerator must be zero. Therefore, solving for the values of \( a \) and \( b \): \[ 1 + a - b = 0 \quad \text{and} \quad 2a + 8a = 0 \Rightarrow a = -\frac{1}{4} \] Now, substitute \( a \) into the equation: \[ b = a + 1 \quad \Rightarrow \quad b = -\frac{1}{4} + 1 = \frac{3}{4} \] Thus, we get: \[ a + b = -\frac{1}{4} + \frac{3}{4} = \frac{1}{2} \]