Let $\alpha$ and $\beta$ be the real numbers such that
$$\lim_{x \to 0} \dfrac{1}{x^3}\left(\dfrac{\alpha}{2}\int_0^x \dfrac{1}{1-t^2}\,dt + \beta x\cos x\right) = 2.$$
Then the value of $\alpha + \beta$ is ___.
Step-by-Step Solution
Key Concept: L'Hôpital via Taylor expansion; matching coefficients for limit to exist
Expand using Taylor series near $x=0$.
$\dfrac{1}{1-t^2} = 1 + t^2 + t^4 + \cdots$, so $\int_0^x \dfrac{1}{1-t^2}\,dt = x + \dfrac{x^3}{3} + \dfrac{x^5}{5} + \cdots$
$\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$, so $x\cos x = x - \dfrac{x^3}{2} + \cdots$
Numerator: $\dfrac{\alpha}{2}\left(x + \dfrac{x^3}{3} + \cdots\right) + \beta\left(x - \dfrac{x^3}{2} + \cdots\right)$
$= \left(\dfrac{\alpha}{2}+\beta\right)x + \left(\dfrac{\alpha}{6} - \dfrac{\beta}{2}\right)x^3 + \cdots$
For the limit $\dfrac{\text{numerator}}{x^3}$ to exist and equal 2, the coefficient of $x$ must vanish:
$\dfrac{\alpha}{2} + \beta = 0 \Rightarrow \alpha = -2\beta$.
Coefficient of $x^3$: $\dfrac{\alpha}{6} - \dfrac{\beta}{2} = 2$.
Substituting $\alpha = -2\beta$: $\dfrac{-2\beta}{6} - \dfrac{\beta}{2} = -\dfrac{\beta}{3} - \dfrac{\beta}{2} = -\dfrac{5\beta}{6} = 2 \Rightarrow \beta = -\dfrac{12}{5}$.
$\alpha = \dfrac{24}{5}$.
$\alpha + \beta = \dfrac{24}{5} - \dfrac{12}{5} = \dfrac{12}{5} = 2.4$.
Correct Answer: 2.4