$ \lim _{x \rightarrow 0} \frac{1}{x} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) $ is equal to

Question

$ \lim _{x \rightarrow 0} \frac{1}{x} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) $ is equal to, $ -2 $, $ 0 $, $ 2 $, $ \infty $

MARKS App · Accepted Answer

Put x = tan ⁡ θ ⇒ θ = tan - 1 ⁡ x As x → 0 ⇒ θ → 0 ∴ lim θ → 0 ⁡ 1 tan ⁡ θ sin - 1 ⁡ 2 tan ⁡ θ 1 + tan 2 ⁡ θ = l i m θ → 0 1 tan ⁡ θ sin - 1 ⁡ ( sin ⁡ 2 θ ) = l i m θ → 0 2 θ tan ⁡ θ = 2

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