Question

$\underset{x\to \frac{1}{\sqrt{2}}}{\lim }\frac{\sin \left({\cos }^{-1}x\right)-x}{1-\tan \left({\cos }^{-1}x\right)}$ is equal to :

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $-\frac{1}{\sqrt{2}}$

Answer 1

Answer: (D)

$\underset{x\to \frac{1}{\sqrt{2}}}{\lim }\frac{\sin \left({\cos }^{-1}x\right)-x}{1-\tan \left({\cos }^{-1}x\right)}$ $\underset{x\to \frac{1}{\sqrt{2}}}{\lim }\frac{\sin \left({\sin }^{-1}\sqrt{1-x^2}\right)-x}{1-\tan \left({\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right)}$
$\underset{x\to \frac{1}{\sqrt{2}}}{\lim }\frac{\sqrt{1-x^2}-x}{1-\left(\frac{\sqrt{1-x^2}}{x}\right)}$ $\underset{x\to \frac{1}{\sqrt{2}}}{\lim }(-x)=-\frac{1}{\sqrt{2}}$