Question 11

We have $\int_0^x \frac{1}{1-t^2} dt = \frac{1}{2} \ln \left| \frac{1+x}{1-x} \right| = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$ Substitute the expansion into the limit: $\lim_{x \to 0} \frac{\frac{\alpha}{2}(x + \frac{x^3}{3}) + \beta x (1 - \frac{x^2}{2})}{x^3} = 2$ $\lim_{x \to 0} \frac{(\frac{\alpha}{2} + \beta)x + (\frac{\alpha}{6} - \frac{\beta}{2})x^3}{x^3} = 2$ For the limit to exist and be finite, the coefficient of $x$ must be zero: $\frac{\alpha}{2} + \beta = 0 \implies \alpha = -2\beta$ The limit then becomes the coefficient of $x^3$: $\frac{\alpha}{6} - \frac{\beta}{2} = 2$ Substitute $\alpha = -2\beta$: $\frac{-2\beta}{6} - \frac{\beta}{2} = 2 \implies -\frac{\beta}{3} - \frac{\beta}{2} = 2 \implies -\frac{5\beta}{6} = 2 \implies \beta = -2.4$ Then $\alpha = -2(-2.4) = 4.8$. The value of $\alpha + \beta = 4.8 - 2.4 = 2.4$.