Question

Let $ \alpha $ and $ \beta $ be the real numbers such that $$ \lim_{x \to 0} \frac{1}{x^3} \left( \frac{\alpha}{2} \int_{0}^{x} \frac{1}{1 - t^2} \, dt + \beta x \cos x \right) = 2. $$ Then the value of $ \alpha + \beta $ is __________.

Hint:

In limit problems involving integrals, expand the integral using a Taylor series if the integrand is analytic and simplify before applying the limit.

Answer 1

Step 1: Expand the integral near \( x = 0 \) We have: \[ \int_0^x \frac{1}{1 - t^2} dt \] Use the Taylor expansion: \[ \frac{1}{1 - t^2} = 1 + t^2 + t^4 + \dots \Rightarrow \int_0^x \frac{1}{1 - t^2} dt = \int_0^x \left( 1 + t^2 + t^4 + \dots \right) dt \] \[ = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \] So, \[ \frac{\alpha}{2} \int_0^x \frac{1}{1 - t^2} dt = \frac{\alpha}{2} \left( x + \frac{x^3}{3} + \dots \right) = \frac{\alpha x}{2} + \frac{\alpha x^3}{6} + \dots \] Step 2: Expand \( \beta x \cos x \) \[ \beta x \cos x = \beta x (1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots) = \beta x - \frac{\beta x^3}{2} + \dots \] Step 3: Combine expressions inside the limit \[ \frac{1}{x^3} \left( \frac{\alpha x}{2} + \frac{\alpha x^3}{6} + \beta x - \frac{\beta x^3}{2} + \dots \right) = \frac{1}{x^3} \left( \left( \frac{\alpha}{2} + \beta \right)x + \left( \frac{\alpha}{6} - \frac{\beta}{2} \right)x^3 + \dots \right) \] Step 4: Apply the limit For the limit to be finite, the coefficient of \( \frac{1}{x^2} \) must vanish: \[ \frac{\alpha}{2} + \beta = 0 \quad \Rightarrow \quad \beta = -\frac{\alpha}{2} \] Now substitute: \[ \text{Limit} = \lim_{x \to 0} \frac{1}{x^3} \left( \left( \frac{\alpha}{6} - \frac{\beta}{2} \right)x^3 \right) = \frac{\alpha}{6} - \frac{\beta}{2} = 2 \] Now plug in \( \beta = -\frac{\alpha}{2} \): \[ \frac{\alpha}{6} - \left(-\frac{\alpha}{4}\right) = \frac{\alpha}{6} + \frac{\alpha}{4} = \frac{2\alpha + 3\alpha}{12} = \frac{5\alpha}{12} \Rightarrow \frac{5\alpha}{12} = 2 \Rightarrow \alpha = \frac{24}{5} \Rightarrow \beta = -\frac{12}{5} \] Final Answer: \[ \alpha + \beta = \frac{24}{5} - \frac{12}{5} = \frac{12}{5} \]