Question

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

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

Answer 1

Answer: (A)

Let ${\sin }^{-1}x=\theta$.
Then $x=\sin \theta$.
Now, $x\to \frac{1}{\sqrt{2}}$
$\Rightarrow \sin \theta \to \frac{1}{\sqrt{2}}$
$\Rightarrow \theta \to \frac{\pi }{4}$
$\therefore \underset{x\to \frac{1}{\sqrt{2}}}{\lim }\frac{x-\cos \left({\sin }^{-1}x\right)}{1-\tan \left({\sin }^{-1}x\right)}$
$=\underset{\theta \to \frac{\pi }{4}}{\lim }\frac{\sin \theta -\cos \theta }{1-\tan \theta }$
$=\underset{\theta \to \frac{\pi }{4}}{\lim }\frac{(\sin \theta -\cos \theta )}{(\cos \theta -\sin \theta )}\cos \theta$
$=\underset{\theta \to \frac{\pi }{4}}{\lim }-\cos \theta =-\frac{1}{\sqrt{2}}$