Let $$f : \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$$ be a differentiable function such that $$f(0) = \frac{1}{2}$$. If $$\lim_{x \to 0} \frac{x \int_0^x f(t) dt}{e^{x^2} - 1} = \alpha$$, then $$8\alpha^2$$ is equal to :

 Notice the limit is in $$\frac{0}{0}$$ form.

 Use the standard limit $$\lim_{x \to 0} \frac{e^{x^2}-1}{x^2} = 1$$.

 Rewrite the expression: $$\alpha = \lim_{x \to 0} \frac{x \int_0^x f(t)dt}{x^2} = \lim_{x \to 0} \frac{\int_0^x f(t)dt}{x}$$.

Apply L'Hôpital's Rule (Leibniz Rule for the numerator):

$$\alpha = \lim_{x \to 0} \frac{f(x)}{1} = f(0) = \frac{1}{2}$$

Calculate $$8\alpha^2 = 8(1/2)^2 = 8(1/4) = 2$$.