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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
(1) $\frac{1}{6}$
(2) $\frac{1}{2}$
(3) $6$
(4) $2$
$\lim_{x\to 0} \frac{\left(x + \frac{x^3}{6} + \cdots\right) - \left(x - \frac{x^3}{3} + \cdots\right)}{3x^3} = \frac{1}{6}$. So $6L + 1 = 6 \cdot \frac{1}{6} + 1 = 2$.
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