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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, $$ \begin{aligned} \lim_{x\to+\infty}f'(x) &=\lim_{x\to+\infty}\frac{2g'(x)}{\sin2g(x)} \\ &=\lim_{x\to+\infty}\frac{\pi/\cosh^2x}{\pi(1-\tanh x)} \\ &=\lim_{x\to+\infty}\frac{1}{(\cosh x-\sinh x)\cosh x} \\ &=\lim_{x\to+\infty}\frac{2}{e^{-x}(e^x+e^{-x})} \\ &=\lim_{x\to+\infty}\frac{2}{1+e^{-2x}} =2. \end{aligned} $$