Question

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

• A. $1$
• B. $0$
• C. $2$
• D. None of these

Answer 1

Answer: (D)

None of these

We know that
${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\{\begin{array}{cc}2ta{n}^{-1}x,& x\ge 0\\ -2{\tan }^{-1}x,& x<0\end{array}$
$RHL=\underset{x\to 0^{+}}{lim}\frac{1}{x}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\underset{x\to 0^{+}}{lim}\frac{2{\tan }^{-1}x}{x}=2$
$LHL=\underset{x\to 0^{-}}{lim}\frac{1}{x}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\underset{x\to 0^{-}}{lim}\frac{-2{\tan }^{-1}x}{x}=-2$
Hence, $LHL\ne RHL$
So, the limit does not exist.