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

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Limits

lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}

is equal to

L

, then the value of

(6 L + 1)

is :

Answer: (C)

Near

x = 0

sin^{- 1} x = x + \frac{x ^{3}}{6} + O (x^{5}), tan^{- 1} x = x - \frac{x ^{3}}{3} + O (x^{5})

so the numerator is

\frac{x ^{3}}{6} + \frac{x ^{3}}{3} + O (x^{5}) = \frac{x ^{3}}{2} + O (x^{5})

. Dividing by

3 x^{3}

and letting

x \to 0

L = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}

6 L + 1 = 1 + 1 = 2

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option

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