Question

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

Answer 1

$\begin{array}{c}\underset{x\to 1/\sqrt{2}}{lim}\frac{x-\cos \left({\sin }^{-1}x\right)}{1-\tan \left({\sin }^{-1}x\right)}\\ =\underset{x\to 1/\sqrt{2}}{lim}\frac{x-\cos \left({\sin }^{-1}x\right)}{1-\frac{\sin \left({\sin }^{-1}x\right)}{\cos \left({\sin }^{-1}x\right)}}\\ =\underset{x\to 1/\sqrt{2}}{lim}\frac{x-\cos \left({\sin }^{-1}x\right)}{1-\frac{x}{\cos \left({\sin }^{-1}x\right)}}\\ =\underset{x\to 1/\sqrt{2}}{lim}\frac{x-\cos \left({\sin }^{-1}x\right)}{\frac{\cos \left({\sin }^{-1}x\right)-x}{\cos \left({\sin }^{-1}x\right)}}\\ =\underset{x\to 1/\sqrt{2}}{lim}-\cos \left({\sin }^{-1}x\right)\\ =-\cos {45}^0=-\frac{1}{\sqrt{2}}\end{array}$