The value of $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(x-[x]^{2}\right) \cdot \sin ^{-1}\left(x-[x]^{2}\right)}{x-x^{3}}$, where $[x]$ denotes the greatest integer $\leq x$ is :

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Limits

The value of

lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}

, where

[x]

denotes the greatest integer

\leq x

is :

Answer: (C)

x \to 0^{+}

, we have

[x] = 0

(greatest integer function). Thus

x - [x]^{2} = x - 0 = x

. The limit becomes

lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x ) \cdot s i n ^{- 1} ( x )}{x - x ^{3}} = lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x ) \cdot s i n ^{- 1} ( x )}{x ( 1 - x ^{2} )}

. As

x \to 0^{+}

cos^{- 1} (x) \to \frac{π}{2}

sin^{- 1} (x) \sim x

, and

1 - x^{2} \to 1

. Therefore the limit is

\frac{\frac{π}{2} \cdot x}{x \cdot 1} = \frac{π}{2}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Single correct option

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