$\lim _{x ightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2} ight)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2} ight) an ^{-1}\left(\frac{3 x-x^3}{1-3 x^2} ight)}$ is equal to

Question

$\lim _{x ightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}ight)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight) 	an ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}ight)}$ is equal to

Aicharya · Accepted Answer

$ \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) \cdot 	an ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}ight)} $ $\because$ We know that $x ightarrow 0^{+}$ $ \begin{aligned} & 2 	an ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^2}=\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) \end{aligned} $ $	herefore$ Limit becomes $ \lim _{x ightarrow 0^{+}} \frac{x \cdot\left(2 	an ^{-1} xight)}{\left(2 	an ^{-1} xight)\left(3 	an ^{-1} xight)} $ $ \Rightarrow \lim _{x ightarrow 0^{+}} \frac{x}{3 	an ^{-1} x} $ Using L Hospital rule $ \lim _{x ightarrow 0^{+}}=\frac{1}{3 	imes \frac{1}{1+x^2}}=\frac{1}{3 	imes\left(\frac{1}{1+0}ight)}=\frac{1}{3} $

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

Solution