Let $$f(x) = \sin^{-1}x$$ and $$g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$$. If $$g(2) = \lim_{x \to 2} g(x)$$, then the domain of the function $$fog$$ is

We have $$g(x) = \dfrac{x^2 - x - 2}{2x^2 - x - 6} = \dfrac{(x-2)(x+1)}{(2x+3)(x-2)}$$. For $$x \neq 2$$, this simplifies to $$g(x) = \dfrac{x + 1}{2x + 3}$$.

Since $$g(2) = \displaystyle\lim_{x \to 2} g(x) = \dfrac{3}{7}$$, the function $$g$$ is defined as $$g(x) = \dfrac{x+1}{2x+3}$$ for all $$x \neq -\dfrac{3}{2}$$.

For $$f \circ g$$ to be defined, we need $$-1 \leq g(x) \leq 1$$ (the domain of $$\sin^{-1}$$).

Condition 1: $$g(x) \geq -1$$, i.e., $$\dfrac{x+1}{2x+3} \geq -1$$. This gives $$\dfrac{x + 1 + 2x + 3}{2x+3} = \dfrac{3x + 4}{2x + 3} \geq 0$$. By sign analysis, this holds when $$x \leq -\dfrac{3}{2}$$ or $$x \geq -\dfrac{4}{3}$$.

Condition 2: $$g(x) \leq 1$$, i.e., $$\dfrac{x+1}{2x+3} \leq 1$$. This gives $$\dfrac{x + 1 - 2x - 3}{2x + 3} = \dfrac{-(x+2)}{2x+3} \leq 0$$, equivalently $$\dfrac{x + 2}{2x + 3} \geq 0$$. By sign analysis, this holds when $$x \leq -2$$ or $$x > -\dfrac{3}{2}$$.

Taking the intersection (and excluding $$x = -\dfrac{3}{2}$$ where $$g$$ is undefined): $$\left(x \leq -\dfrac{3}{2} \text{ or } x \geq -\dfrac{4}{3}\right) \cap \left(x \leq -2 \text{ or } x > -\dfrac{3}{2}\right) = (-\infty, -2] \cup \left[-\dfrac{4}{3}, \infty\right)$$.

Therefore the domain of $$f \circ g$$ is $$\left(-\infty, -2\right] \cup \left[-\dfrac{4}{3}, \infty\right)$$.