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

The expansion of the terms in the given expression begins. The Taylor series expansion of \( \cos x \) around \( x = 0 \) yields:\[\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + O(x^6)\]\[\cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} + O(x^6)\]Consequently, the given expression transforms to:\[L = \lim_{x \to 0} \frac{ \left( 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} \right) + a \left( 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} \right) - b }{x^4}\]Simplification of the expression results in:\[L = \frac{1 + a - b}{x^4} + \frac{a \cdot \frac{(4x)^2}{2!} + O(x^4)}{x^4} + \text{higher order terms}\]For the limit to remain finite, the \( x^4 \) coefficient in the numerator must be zero. Therefore, solving for \( a \) and \( b \):\[1 + a - b = 0 \quad \text{and} \quad 2a + 8a = 0 \Rightarrow a = -\frac{1}{4}\]Substituting \( a \) into the equation gives:\[b = a + 1 \quad \Rightarrow \quad b = -\frac{1}{4} + 1 = \frac{3}{4}\]This leads to:\[a + b = -\frac{1}{4} + \frac{3}{4} = \frac{1}{2}\]