Question

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

• A. $1$
• B. $0$
• C. $2$
• D. $Doesnotexist$

Answer 1

Answer: (C)

$2$

$\begin{array}{c}\underset{x\to 0}{lim}\frac{{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)}{x}\\ put,x=\tan \theta \\ x\to 0\Rightarrow \theta \to 0\\ \underset{\theta \to 0}{lim}\frac{{\cos }^{-1}\left(\frac{1-\tan {\theta }^2}{1+\tan {\theta }^2}\right)}{\tan \theta }\\ \underset{\theta \to 0}{lim}\frac{{\cos }^{-1}\left(\cos 2\theta \right)}{\tan \theta }\\ \underset{\theta \to 0}{lim}\frac{2\theta }{\tan \theta }=2\end{array}$