Question

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

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{-1}{\sqrt{2}}$
• C. $0$
• D. $1$

Answer 1

The correct option is B $\frac{-1}{\sqrt{2}}$

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

Applying limit

$\frac{\frac{1}{\sqrt{2}}-cos\left(si{n}^{-1}\frac{1}{\sqrt{2}}\right)}{\frac{1}{\sqrt{2}}-tan\left(si{n}^{-1}\frac{1}{\sqrt{2}}\right)}$

$=\frac{\frac{1}{\sqrt{2}}-cos\left(\frac{\pi }{4}\right)}{\frac{1}{\sqrt{2}}-tan\left(\frac{\pi }{4}\right)}$

$=\frac{\frac{1}{\sqrt{2}}-cos\left(\frac{\pi }{4}\right)}{\frac{1}{\sqrt{2}}-tan\left(\frac{\pi }{4}\right)}$

$=\frac{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-1}$

$=0$