lim limits_{x} {rightarrow frac{1}{sqrt{2}}} | vedclass

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(D) Let $f(x) = \frac{\sin(\cos^{-1} x) - x}{1 - 	an(\cos^{-1} x)}$.We know that $\cos^{-1} x = \sin^{-1}(\sqrt{1-x^2})$ and $\cos^{-1} x = 	an^{-1}

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

Vedclass · Answer

(D) Let $f(x) = \frac{\sin(\cos^{-1} x) - x}{1 - 	an(\cos^{-1} x)}$.We know that $\cos^{-1} x = \sin^{-1}(\sqrt{1-x^2})$ and $\cos^{-1} x = 	an^{-1}(\frac{\sqrt{1-x^2}}{x})$.Substituting these into the expression:$\lim \limits_{x}$ ${ightarrow \frac{1}{\sqrt{2}}} \frac{\sin(\sin^{-1}(\sqrt{1-x^2})) - x}{1 - 	an(	an^{-1}(\frac{\sqrt{1-x^2}}{x}))}$$= \lim \limits_{x ightarrow \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2} - x}{1 - \frac{\sqrt{1-x^2}}{x}}$$= \lim \limits_{x ightarrow \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2} - x}{\frac{x - \sqrt{1-x^2}}{x}}$$= \lim \limits_{x ightarrow \frac{1}{\sqrt{2}}} \frac{- (x - \sqrt{1-x^2})}{\frac{x - \sqrt{1-x^2}}{x}}$$= \lim \limits_{x ightarrow \frac{1}{\sqrt{2}}} (-x) = -\frac{1}{\sqrt{2}}$.

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

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