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. None of these

Answer 1

Answer: (D)

We know that ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\{\begin{array}{ll}2{\tan }^{-1}x,& x\ge 0\\ -2{\tan }^{-1}x,& x\le 0\end{array}$
or $\underset{x\to 0^{+}}{\lim }\frac{1}{x}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$=\underset{x\to 0^{+}}{\lim }\frac{2{\tan }^{-1}x}{x}=2$
and $\underset{x\to 0^{-}}{\lim }\frac{1}{x}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$=\underset{x\to 0^{+}}{\lim }\left[-\frac{2{\tan }^{-1}x}{x}\right]=-2$