If cot[f(x)] = frac{3x - x^3}{1 - 3x^2} and s | vedclass

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(B) Given $\cot[f(x)] = \frac{3x - x^3}{1 - 3x^2} = 	an(3	heta)$ where $x = 	an 	heta$. Thus, $f(x) = 	ext{arccot}(	an(3	heta)) = \frac{\pi}{2}

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$\frac{3}{2}$

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(B) Given $\cot[f(x)] = \frac{3x - x^3}{1 - 3x^2} = 	an(3	heta)$ where $x = 	an 	heta$. Thus, $f(x) = 	ext{arccot}(	an(3	heta)) = \frac{\pi}{2} - 3\arctan(x)$.Then $f'(x) = -\frac{3}{1+x^2}$.Given $\sin[g(x)] = \frac{1 - x^2}{1 + x^2} = \cos(2	heta)$ where $x = 	an 	heta$. Thus, $g(x) = \arcsin(\cos(2	heta)) = \arcsin(\sin(\frac{\pi}{2} - 2	heta)) = \frac{\pi}{2} - 2\arctan(x)$.Then $g'(x) = -\frac{2}{1+x^2}$.Using $L$'$H$ôpital's rule, $\lim_{x 	o t} \frac{f(x) - f(t)}{g(x) - g(t)} = \frac{f'(t)}{g'(t)} = \frac{-3/(1+t^2)}{-2/(1+t^2)} = \frac{3}{2}$.

If $\cot[f(x)] = \frac{3x - x^3}{1 - 3x^2}$ and $\sin[g(x)] = \frac{1 - x^2}{1 + x^2}$, then $\lim_{x \to t} \frac{f(x) - f(t)}{g(x) - g(t)} = \dots$

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