Idea for $\lim\limits_{x \to \frac{\pi}{2}} \left( \tan \left( \tfrac{\pi}{4} \sin x\right)\right)^{1/ ( \tan (\pi \sin x))}$ $$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.
ps. can't use lhopital
 A: Warning: you'll want to double-check all my arithmetic.
With $y:=\pi\sin x$ we can rewrite this as $\exp\lim_{y\to\pi}\frac{\ln\tan\frac{y}{4}}{\tan y}$. Define $t:=\tan\frac{y}{4}$ so $$\tan\frac{y}{2}=\frac{2t}{1-t^2},\,\tan y=\frac{4t(1-t^2)}{1-6t^2+t^4}.$$Then your limit is $$\exp\lim_{t\to 1}\frac{(1-6t^2+t^4)\ln t}{4t(1-t^2)}=\exp-\lim_{t\to 1}\frac{\ln t}{t(1-t^2)}.$$Finally, write $t=1+\epsilon$ so the limit is$$\exp\lim_{\epsilon\to 0}\frac{\ln(1+\epsilon)}{(1+\epsilon)\epsilon(2+\epsilon)}=\exp\lim_{\epsilon\to 0}\frac{\ln(1+\epsilon)}{2\epsilon}=\sqrt{e}.$$
A: You might first do the substitution $t=\frac{\pi}{4}\sin x$ that brings the limit into the form
$$
\lim_{t\to\pi/4^-}(\tan t)^{1/\tan4t}
$$
(prove first that both the limit from the left and from the right lead to the same limit).
Now it's better to pass to the logarithm:
$$
\lim_{t\to\pi/4^-}\frac{\log\tan t}{\tan 4t}
$$
Apply $t=\pi/4-u$,
$$
\tan t=\frac{1-\tan u}{1+\tan u}
$$
and $\tan4t=\tan(\pi-4u)=-\tan 4u$. Therefore the limit becomes
$$
\lim_{u\to0^+}\frac{\log(1+\tan u)-\log(1-\tan u)}{\tan 4u}
$$
Now, for instance,
$$
\lim_{u\to0^+}\frac{\log(1+\tan u)}{\tan4u}=
\lim_{u\to0^+}\frac{\log(1+\tan u)}{\tan u}\frac{\tan u}{u}\frac{4u}{\tan 4u}\frac{1}{4}
$$
A: Alternatively:
$$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=\\
\lim_{x \to \frac{\pi}{2}} \left[1+\tan \left(\frac{\pi}{4}\sin x\right)-1\right]^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{4}\sin x\right)-1}{\tan(\pi \sin x)} \right]=\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}\sin x\right)-\cos \left(\frac{\pi}{4}\sin x\right)}{\sin(\pi \sin x)}\cdot \frac{\cos (\pi \sin x)}{\cos \left(\frac{\pi}{4}\sin x\right)}\right] =\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}\sin x\right)}{\sin(\pi \sin x)}\right] =\\
\exp \left[\lim_{x \to \frac{\pi}{2}} \frac{1}{2\sin(\frac{\pi}{2} \sin x)}\right] =\exp \left[\frac12\right]=\sqrt{e}.
$$
A: Let $\dfrac{\pi\sin x}4=y$
$$\lim_{y\to\frac\pi4}(\tan y)^{\frac1{\tan4y}}=\left[\lim_{y\to\frac\pi4}(1+\tan y-1)^{\frac1{\tan y -1}}\right]^{\lim_{y\to\frac\pi4}\dfrac{\tan y-1}{\tan4y}}$$
The inner limit converges to $e$
The exponent$(F)= \lim_{y\to\frac\pi4}\dfrac{\tan y-1}{\tan4y}=\lim_{y\to\frac\pi4}\dfrac{\tan y-\tan\dfrac\pi4}{\tan4y-\tan\pi}$
Method$\#1:$
$$F=\dfrac{\dfrac{d(\tan y)}{dy}_{(\text { at }y=\pi/4)}}{\dfrac{d(\tan4y)}{dy}_{(\text { at }y=\pi/4)}}=?$$
Method$\#2:$
$$F=\lim_{y\to\frac\pi4}\dfrac{\sin\left(y-\dfrac\pi4\right)\cos4y}{\sin4y\cos y\cos\dfrac\pi4}$$
$$=\sqrt2\lim_{y\to\frac\pi4}\dfrac{\cos4y}{\cos y}\cdot\dfrac{\lim_{y\to\frac\pi4}\dfrac{\sin\left(y-\dfrac\pi4\right)}{\left(y-\dfrac\pi4\right)}}{-4\lim_{y\to\frac\pi4}\dfrac{\sin4\left(y-\dfrac\pi4\right)}{4\left(y-\dfrac\pi4\right)}}\text{ as }\sin(4y-\pi)=-\sin(\pi-4y)=-\sin4y$$
$$=\dfrac{\sqrt2\cos\pi}{-4\cos\dfrac\pi4}=?$$
A: Using trigonometric identities, by $y=\frac \pi 2 \sin x \to \frac \pi 2$ and $t=\frac \pi 2 -y \to 0$, we obtain:


*

*$\tan \frac{\theta}{2}  = \frac{\sin \theta}{1 +
   \cos \theta} \implies \tan \left(\frac{\pi}{4}\sin x\right)=\frac{\sin y}{1 +
   \cos y}=\frac{\cos t}{1 +
   \sin t}$

*$\tan (2\theta) = \frac{2 \tan \theta} {1 - \tan^2 \theta} \implies \tan(\pi \sin x)=\frac{2 \tan y} {1 - \tan^2 y}=\frac{2 \cot t} {1 - \cot^2 t}=\frac{2\cos t \sin t}{\sin^2 t-\cos^2t}$
then
$$\tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right)=e^{{\dfrac {\log \left(\tan \left(\frac{\pi}{4}\sin x\right)\right)}{\tan(\pi \sin x)} }}
=e^{\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\log\left(\frac{\cos t}{1 +
   \sin t} \right)}\to e^\frac12$$
indeed
$$\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot \log\left(\frac{\cos t}{1 +
   \sin t} \right)=\\=\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot\frac{\cos t-\sin t-1}{1+\sin t}\cdot\log\left(1+\frac{\cos t-\sin t-1}{1 +
   \sin t} \right)^{\frac{1+\sin t}{\cos t-\sin t -1}}\to\frac 12 \cdot \log e=\frac 12$$
since
$$\frac{\sin^2 t-\cos^2t}{2\cos t \sin t}\cdot\frac{\cos t-\sin t-1}{1+\sin t}=\frac{\cos^2t-\sin^2 t}{\cos t(1+\sin t)}\cdot\frac{\sin t+1-\cos t}{2\sin t}=1\cdot \frac12$$
and by standard limits
$$\frac{\sin t+1-\cos t}{2\sin t}=\frac{ \frac{\sin t}{t}+t\frac{1-\cos t}{t^2} } {2\frac {\sin t} t}\to \frac{1+0\cdot 1}{2\cdot 1}=\frac12$$
