If $$\lim_{x \to 0} \frac{\sin^{-1}x - \tan^{-1}x}{3x^3}$$ is equal to $$L$$, then the value of $$(6L + 1)$$ is:

We use Taylor series expansions around $$x = 0$$. We have $$\sin^{-1}x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \ldots$$ and $$\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \ldots$$

So $$\sin^{-1}x - \tan^{-1}x = \left(\frac{1}{6} + \frac{1}{3}\right)x^3 + \ldots = \frac{1}{2}x^3 + \ldots$$

Therefore $$L = \lim_{x \to 0} \frac{\sin^{-1}x - \tan^{-1}x}{3x^3} = \frac{1/2}{3} = \frac{1}{6}$$.

Hence $$(6L + 1) = 6 \cdot \frac{1}{6} + 1 = 1 + 1 = 2$$.

The answer is Option D: $$2$$.