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Let $$f(x)=\log(\tan g(x)),$$ with $$g(x)=\frac{\pi}{2}\tanh x$$ It can be seen that $f$ is a differentiable function in $(0,+\infty)$, so it makes sense to try to compute $$\lim_{x\to+\infty}f'(x).$$ To this end, we first obtain $f'$ by applying the chain rule: $$f'(x)=\frac{1}{\tan g(x)} \frac{1}{\cos^2 g(x)}g'(x),$$ where $$g'(x)=\frac{\pi}{2}\frac{1}{\cosh^2 x}.$$ By using some trigonometric relations, we get $$\tan g(x)\,\cos^2 g(x)=\sin g(x)\cos g(x)=\frac{1}{2}\sin 2g(x).$$ Hence, $$f'(x)=\frac{2g'(x)}{\sin2g(x)}.$$ SageMath can handle the limit of the above expression of $f'$:

sage: g(x) = (pi/2) * tanh(x)                                                   
sage: gp = g.derivative()                                                       
sage: limit(2*gp(x)/sin(2*g(x)), x=+oo)                                         
2

Consequently, $$\lim_{x\to+\infty}f'(x)=2,$$ as conjectured by @Emmanuel Charpentier in his comment.

It is not hard to finish the computation by hand. Since $$\lim_{x\to+\infty}2g(x)=\pi \lim_{x\to+\infty}\tanh x=\pi,$$ we have the following equivalence as $x\to+\infty$: $$\sin 2g(x)=\sin(\pi-2g(x))=\sin\pi(1-\tanh x)\sim \pi(1-\tanh x).$$ Therefore, $$\begin{aligned} \lim_{x\to+\infty}f'(x) &=\lim_{x\to+\infty}\frac{2g'(x)}{\sin2g(x)} \\ &=\lim_{x\to+\infty}\frac{\pi/\cosh^2x}{\pi(1-\tanh x)} \\ &=\lim_{x\to+\infty}\frac{1}{(\cosh x-\sinh x)\cosh x} \\ &=\lim_{x\to+\infty}\frac{2}{e^{-x}(e^x+e^{-x})} \\ &=\lim_{x\to+\infty}\frac{2}{1+e^{-2x}} =2. \end{aligned}$$

Let $$f(x)=\log(\tan g(x)),$$ with $$ g(x)=\frac{\pi}{2}\tanh x x. $$It can be seen that $f$ is a differentiable function in $(0,+\infty)$, so it makes sense to try to compute$$ \lim_{x\to+\infty}f'(x). $$To this end, we first obtain $f'$ by applying the chain rule:$$ f'(x)=\frac{1}{\tan g(x)} \frac{1}{\cos^2 g(x)}g'(x), $$where$$ g'(x)=\frac{\pi}{2}\frac{1}{\cosh^2 x}. $$By using some trigonometric relations, we get$$ \tan g(x)\,\cos^2 g(x)=\sin g(x)\cos g(x)=\frac{1}{2}\sin 2g(x). $$Hence,$$ f'(x)=\frac{2g'(x)}{\sin2g(x)}. $$ SageMath can handle the limit of the above expression of $f'$:

sage: g(x) = (pi/2) * tanh(x)                                                   
sage: gp = g.derivative()                                                       
sage: limit(2*gp(x)/sin(2*g(x)), x=+oo)                                         
2

Consequently, $$\lim_{x\to+\infty}f'(x)=2,$$ as shown in part of the results shown in the opening post or conjectured by @Emmanuel Charpentier in his comment.

It is not hard to finish the computation by hand. Since $$\lim_{x\to+\infty}2g(x)=\pi \lim_{x\to+\infty}\tanh x=\pi,$$ we have the following equivalence as $x\to+\infty$: $$\sin 2g(x)=\sin(\pi-2g(x))=\sin\pi(1-\tanh x)\sim \pi(1-\tanh x).$$ Therefore, $$\begin{aligned} \lim_{x\to+\infty}f'(x) &=\lim_{x\to+\infty}\frac{2g'(x)}{\sin2g(x)} \\ &=\lim_{x\to+\infty}\frac{\pi/\cosh^2x}{\pi(1-\tanh x)} \\ &=\lim_{x\to+\infty}\frac{1}{(\cosh x-\sinh x)\cosh x} \\ &=\lim_{x\to+\infty}\frac{2}{e^{-x}(e^x+e^{-x})} \\ &=\lim_{x\to+\infty}\frac{2}{1+e^{-2x}} =2. \end{aligned}$$

Let $$f(x)=\log(\tan g(x)),$$ with $$g(x)=\frac{\pi}{2}\tanh x.$$ It can be seen that $f$ is a differentiable function in $(0,+\infty)$, so it makes sense to try to compute $$\lim_{x\to+\infty}f'(x).$$ To this end, we first obtain $f'$ by applying the chain rule: $$f'(x)=\frac{1}{\tan g(x)} \frac{1}{\cos^2 g(x)}g'(x),$$ where $$g'(x)=\frac{\pi}{2}\frac{1}{\cosh^2 x}.$$ By using some trigonometric relations, we get $$\tan g(x)\,\cos^2 g(x)=\sin g(x)\cos g(x)=\frac{1}{2}\sin 2g(x).$$ Hence, $$f'(x)=\frac{2g'(x)}{\sin2g(x)}.$$ SageMath can handle the limit of the above expression of $f'$:

sage: g(x) = (pi/2) * tanh(x)                                                   
sage: gp = g.derivative()                                                       
sage: limit(2*gp(x)/sin(2*g(x)), x=+oo)                                         
2

Consequently, $$\lim_{x\to+\infty}f'(x)=2,$$ as shown in part of the results shown reported in the opening post or conjectured by @Emmanuel Charpentier in his comment.

It is not hard to finish the computation by hand. Since $$\lim_{x\to+\infty}2g(x)=\pi \lim_{x\to+\infty}\tanh x=\pi,$$ we have the following equivalence as $x\to+\infty$: $$\sin 2g(x)=\sin(\pi-2g(x))=\sin\pi(1-\tanh x)\sim \pi(1-\tanh x).$$ Therefore, $$\begin{aligned} \lim_{x\to+\infty}f'(x) &=\lim_{x\to+\infty}\frac{2g'(x)}{\sin2g(x)} \\ &=\lim_{x\to+\infty}\frac{\pi/\cosh^2x}{\pi(1-\tanh x)} \\ &=\lim_{x\to+\infty}\frac{1}{(\cosh x-\sinh x)\cosh x} \\ &=\lim_{x\to+\infty}\frac{2}{e^{-x}(e^x+e^{-x})} \\ &=\lim_{x\to+\infty}\frac{2}{1+e^{-2x}} =2. \end{aligned}$$