If _ x → 0 ^ - 1 x - ^ - 1 x 3 x ^ 3 is equal to L, then the value...

Given $\lim_{x \to 0}\dfrac{\sin^{- 1}x - \tan^{- 1}x}{3{x}^{3}}$

$⇒ \lim_{x \to 0}\dfrac{\left(x + \dfrac{{x}^{3}}{3 !} \ldots\right) - \left(x - \dfrac{{x}^{3}}{3} \ldots\right)}{3{x}^{3}}$

$⇒ \lim_{x \to 0}\dfrac{\left(1 + \dfrac{3{x}^{2}}{3 !} \ldots\right) - \left(1 - \dfrac{3{x}^{2}}{3} \ldots\right)}{9{x}^{2}}$ (Use L'Hospital rule $\dfrac{0}{0}$ form)

$⇒ \lim_{x \to 0}\dfrac{\left(\dfrac{6x}{3 !} + \ldots\right) - \left(- \dfrac{6x}{3} + \ldots\right)}{18x}$ (Use L'Hospital rule $\dfrac{0}{0}$ form)

$⇒ \lim_{x \to 0}\dfrac{\left(\dfrac{6}{3 !} + \ldots\right) - \left(- \dfrac{6}{3} + \ldots\right)}{18} = \dfrac{1 + 2}{18} = \dfrac{1}{6}$

Given that, $\lim_{x \to 0}\dfrac{\sin^{- 1}x - \tan^{- 1}x}{3{x}^{3}} = L$

$∴ \dfrac{1}{6} = L ⇒ 6L = 1$

So, $6L + 1 = 2$