Evaluate Real Parameters in a Limit with an Integral

To find the value of $\alpha + \beta$, we evaluate the given limit using Taylor series expansions around $x = 0$.

First, consider the Taylor series expansion of the integrand: $\frac{1}{1-t^2} = 1 + t^2 + O(t^4)$

Integrating this from $0$ to $x$: $\int_0^x \frac{1}{1-t^2}\,dt = \int_0^x \left(1 + t^2 + O(t^4)\right)\,dt = x + \frac{x^3}{3} + O(x^5)$

Next, consider the Taylor series expansion of $\cos x$: $\cos x = 1 - \frac{x^2}{2} + O(x^4)$

Thus, $\beta x \cos x = \beta x \left(1 - \frac{x^2}{2} + O(x^4)\right) = \beta x - \frac{\beta x^3}{2} + O(x^5)$

Now substitute these expansions into the expression inside the limit: $\frac{\alpha}{2}\int_0^x \frac{1}{1-t^2}\,dt + \beta x \cos x = \frac{\alpha}{2}\left(x + \frac{x^3}{3}\right) + \beta\left(x - \frac{x^3}{2}\right) + O(x^5)$ $= \left(\frac{\alpha}{2} + \beta\right)x + \left(\frac{\alpha}{6} - \frac{\beta}{2}\right)x^3 + O(x^5)$

Substituting this back into the limit gives: $\lim_{x \to 0} \frac{1}{x^3}\left( \left(\frac{\alpha}{2} + \beta\right)x + \left(\frac{\alpha}{6} - \frac{\beta}{2}\right)x^3 + O(x^5) \right) = 2$

For this limit to exist and be finite, the coefficient of the linear term in $x$ must be zero: $\frac{\alpha}{2} + \beta = 0 \implies \alpha = -2\beta$

Equating the coefficient of $x^3$ to the value of the limit: $\frac{\alpha}{6} - \frac{\beta}{2} = 2$

Substitute $\alpha = -2\beta$ into this equation: $\frac{-2\beta}{6} - \frac{\beta}{2} = 2$ $-\frac{\beta}{3} - \frac{\beta}{2} = 2$ $-\frac{5\beta}{6} = 2 \implies \beta = -\frac{12}{5} = -2.4$

Then, $\alpha = -2\beta = -2\left(-\frac{12}{5}\right) = \frac{24}{5} = 4.8$

Finally, we calculate $\alpha + \beta$: $\alpha + \beta = 4.8 + (-2.4) = 2.4$