lim _{x rightarrow 0^{+}} frac{x sin ^{-1}lef | vedclass

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Question

(B) We are given the limit: $\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) We are given the limit: $\lim _{x ightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}ight)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight) 	an ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}ight)}$.Since $x ightarrow 0^{+}$,we use the following trigonometric identities:$2 	an ^{-1} x = \sin ^{-1} \left(\frac{2 x}{1+x^2}ight) = \cos ^{-1} \left(\frac{1-x^2}{1+x^2}ight)$$3 	an ^{-1} x = 	an ^{-1} \left(\frac{3 x-x^3}{1-3 x^2}ight)$Substituting these into the limit expression:$\lim _{x ightarrow 0^{+}} \frac{x \cdot (2 	an ^{-1} x)}{(2 	an ^{-1} x) \cdot (3 	an ^{-1} x)} = \lim _{x ightarrow 0^{+}} \frac{x}{3 	an ^{-1} x}$.Using the standard limit $\lim _{x ightarrow 0} \frac{	an ^{-1} x}{x} = 1$,we get:$\lim _{x ightarrow 0^{+}} \frac{1}{3 \left(\frac{	an ^{-1} x}{x}ight)} = \frac{1}{3 	imes 1} = \frac{1}{3}$.

$\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}$ is equal to

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