Define f: R rightarrow R by f(x) = cos(tan^{- | vedclass

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(A) Given $f(x) = \cos(	an^{-1}(\sin(	an^{-1} x)))$.Using $	an^{-1} x = \sin^{-1} \frac{x}{\sqrt{1+x^2}}$,we have $\sin(	an^{-1} x) = \frac{x}{\sq

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$\frac{3}{2 \sqrt{3}}$

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(A) Given $f(x) = \cos(	an^{-1}(\sin(	an^{-1} x)))$.Using $	an^{-1} x = \sin^{-1} \frac{x}{\sqrt{1+x^2}}$,we have $\sin(	an^{-1} x) = \frac{x}{\sqrt{1+x^2}}$.Then ${f(x) = \cos(	an^{-1}(\frac{x}{\sqrt{1+x^2}}}))$.Using $	an^{-1} 	heta = \cos^{-1} \frac{1}{\sqrt{1+	heta^2}}$,we get $f(x) = \cos(\cos^{-1} \frac{1}{\sqrt{1+\frac{x^2}{1+x^2}}}) = \frac{1}{\sqrt{\frac{1+x^2+x^2}{1+x^2}}} = \sqrt{\frac{1+x^2}{1+2x^2}}$.Now,$(f \circ f)(x) = f(f(x)) = \sqrt{\frac{1+(f(x))^2}{1+2(f(x))^2}}$.Substituting $f(x)^2 = \frac{1+x^2}{1+2x^2}$,we get $(f \circ f)(x) = \sqrt{\frac{1+\frac{1+x^2}{1+2x^2}}{1+2(\frac{1+x^2}{1+2x^2})}} = \sqrt{\frac{1+2x^2+1+x^2}{1+2x^2+2+2x^2}} = \sqrt{\frac{2+3x^2}{3+4x^2}}$.Taking the limit as $x ightarrow \infty$,$\lim_{x ightarrow \infty} \sqrt{\frac{2+3x^2}{3+4x^2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} = \frac{3}{2\sqrt{3}}$.

Define $f: R \rightarrow R$ by $f(x) = \cos(\tan^{-1}(\sin(\tan^{-1} x)))$. Then $\lim_{x \rightarrow \infty} (f \circ f)(x)$ is equal to

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