If lim _{x rightarrow 0} frac{sin ^{-1} x - t | vedclass

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(D) We use the Taylor series expansions for $\sin^{-1} x$ and $	an^{-1} x$ near $x = 0$: $\sin^{-1} x = x + \frac{x^3}{6} + O(x^5)$ $	an^{-1} x = x

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$2$

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(D) We use the Taylor series expansions for $\sin^{-1} x$ and $	an^{-1} x$ near $x = 0$: $\sin^{-1} x = x + \frac{x^3}{6} + O(x^5)$ $	an^{-1} x = x - \frac{x^3}{3} + O(x^5)$ Substituting these into the limit expression: $L = \lim_{x ightarrow 0} \frac{(x + \frac{x^3}{6}) - (x - \frac{x^3}{3})}{3x^3}$ $L = \lim_{x ightarrow 0} \frac{\frac{x^3}{6} + \frac{x^3}{3}}{3x^3} = \lim_{x ightarrow 0} \frac{\frac{1}{6} + \frac{1}{3}}{3} = \frac{\frac{1+2}{6}}{3} = \frac{3/6}{3} = \frac{1/2}{3} = \frac{1}{6}$ Given $L = 1/6$,we calculate $6L + 1$: $6(1/6) + 1 = 1 + 1 = 2$

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

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