If x = log_e left( cot left( frac{pi}{4} + th | vedclass

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(B) Given $x = \log_e \left( \cot \left( \frac{\pi}{4} + 	heta ight) ight)$.We know that $(\sinh x)(\cosh x) = \frac{1}{2} \sinh(2x)$.Since $e^x

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$-\frac{1}{2}$

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(B) Given $x = \log_e \left( \cot \left( \frac{\pi}{4} + 	heta ight) ight)$.We know that $(\sinh x)(\cosh x) = \frac{1}{2} \sinh(2x)$.Since $e^x = \cot \left( \frac{\pi}{4} + 	heta ight) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$,we have $e^{2x} = \left( \frac{1 - 	an 	heta}{1 + 	an 	heta} ight)^2$ and $e^{-2x} = \left( \frac{1 + 	an 	heta}{1 - 	an 	heta} ight)^2$.Then $\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2} = \frac{1}{2} \left[ \left( \frac{1 - 	an 	heta}{1 + 	an 	heta} ight)^2 - \left( \frac{1 + 	an 	heta}{1 - 	an 	heta} ight)^2 ight]$.Simplifying the expression inside the bracket: $\frac{(1 - 	an 	heta)^4 - (1 + 	an 	heta)^4}{(1 - 	an^2 	heta)^2} = \frac{-8 	an 	heta - 8 	an^3 	heta}{(1 - 	an^2 	heta)^2} = \frac{-8 	an 	heta (1 + 	an^2 	heta)}{(1 - 	an^2 	heta)^2}$.Thus,$(\sinh x)(\cosh x) = \frac{1}{2} \sinh(2x) = \frac{-4 	an 	heta (1 + 	an^2 	heta)}{(1 - 	an^2 	heta)^2}$.Now,the limit is $\lim_{	heta ightarrow 0} \frac{	heta}{(\sinh x)(\cosh x)} = \lim_{	heta ightarrow 0} \frac{	heta (1 - 	an^2 	heta)^2}{-4 	an 	heta (1 + 	an^2 	heta)}$.Since $\lim_{	heta ightarrow 0} \frac{	heta}{	an 	heta} = 1$,we have $\lim_{	heta ightarrow 0} \frac{1 \cdot (1 - 0)^2}{-4(1 + 0)} = -\frac{1}{4}$.Wait,re-checking the identity: $\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2}$. With $e^x = \frac{1-	an 	heta}{1+	an 	heta}$,$e^{2x} = \frac{1-\sin 2	heta}{1+\sin 2	heta}$.Actually,$\sinh x = \frac{e^x - e^{-x}}{2} = \frac{\cot(\pi/4+	heta) - 	an(\pi/4+	heta)}{2} = -	an(2	heta)/2$ and $\cosh x = \frac{\cot(\pi/4+	heta) + 	an(\pi/4+	heta)}{2} = \frac{1}{\sin(2	heta)}$.So $(\sinh x)(\cosh x) = -\frac{	an(2	heta)}{2 \sin(2	heta)} = -\frac{1}{2 \cos(2	heta)}$.Then $\lim_{	heta ightarrow 0} \frac{	heta}{-1/(2 \cos(2	heta))} = \lim_{	heta ightarrow 0} -2	heta \cos(2	heta) = 0$. Re-evaluating: $x = \log_e(\cot(\pi/4+	heta))$. $\sinh x = \frac{e^x - e^{-x}}{2} = \frac{\cot(\pi/4+	heta) - 	an(\pi/4+	heta)}{2} = \frac{\cos^2(\pi/4+	heta) - \sin^2(\pi/4+	heta)}{2 \sin(\pi/4+	heta)\cos(\pi/4+	heta)} = \frac{\cos(\pi/2+2	heta)}{\sin(\pi/2+2	heta)} = -	an(2	heta)$.$\cosh x = \frac{\cot(\pi/4+	heta) + 	an(\pi/4+	heta)}{2} = \frac{1}{\sin(\pi/2+2	heta)} = \sec(2	heta)$.$(\sinh x)(\cosh x) = -	an(2	heta) \sec(2	heta) = -\frac{\sin(2	heta)}{\cos^2(2	heta)}$.Limit $\lim_{	heta ightarrow 0} \frac{	heta}{-\sin(2	heta)/\cos^2(2	heta)} = \lim_{	heta ightarrow 0} -\frac{	heta}{\sin(2	heta)} \cdot \cos^2(2	heta) = -\frac{1}{2} \cdot 1 = -\frac{1}{2}$.

If $x = \log_e \left( \cot \left( \frac{\pi}{4} + \theta \right) \right)$,then $\lim_{\theta \rightarrow 0} \frac{\theta}{(\sinh x)(\cosh x)} = $

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