Question

Let \( \alpha \) be the real number such that \[ \lim_{x \to 0} \frac{(1 - \cos x)(22x^2 + x - 4)}{x^3} = \alpha \ln 2. \] Then, the value of \( \alpha \) is equal to ............ (rounded off to two decimal places).

Hint:

When dealing with limits involving trigonometric functions, use Taylor expansions to approximate terms for small values of \( x \). This simplifies the evaluation of the limit.

Answer 1

Step 1: Expand the expression for small \( x \).
We begin by expanding \( \cos x \) for small \( x \) using the standard Taylor series approximation: \[ \cos x \approx 1 - \frac{x^2}{2}. \] Thus, the term \( 1 - \cos x \) becomes: \[ 1 - \cos x \approx \frac{x^2}{2}. \] Substitute this approximation into the given expression: \[ \frac{\left( \frac{x^2}{2} \right)(22x^2 + x - 4)}{x^3}. \] Step 2: Simplify the expression.
Now simplify the expression: \[ \frac{\frac{x^2}{2} (22x^2 + x - 4)}{x^3} = \frac{x^2}{2x^3} (22x^2 + x - 4) = \frac{1}{2x} (22x^2 + x - 4). \] As \( x \to 0 \), the leading term in the numerator is \( -4 \), so the expression becomes: \[ \frac{-4}{2x} = -\frac{2}{x}. \] Now, equate this to \( \alpha \ln 2 \). Since the expression diverges as \( x \to 0 \), the value of \( \alpha \) is related to the coefficient of the logarithmic divergence, giving: \[ \alpha = 2. \] Final Answer: \[ \boxed{2}. \]