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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Evaluate: $\tan ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3}) = $ . . . . . . .

Find the value of $\cot \left(\tan ^{-1} a+\cot ^{-1} a\right)$.

If ${x_1}, {x_2}, {x_3}, {x_4}$ are roots of the equation ${x^4} - {x^3}\sin 2\beta + {x^2}\cos 2\beta - x\cos \beta - \sin \beta = 0$,then ${\tan ^{ - 1}}{x_1} + {\tan ^{ - 1}}{x_2} + {\tan ^{ - 1}}{x_3} + {\tan ^{ - 1}}{x_4} = $

The value of $\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]$ is

$\tan ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}\right] + \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module