How to find $\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$? 
Find
  $$\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$$


My attempt: ON THE basis of This post
$$\lim_{x\to1}\tan\frac{\pi x}{4} =1,\quad \lim_{x\to1}\tan\frac{\pi x}{2}=\infty$$
$$\implies\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}= e^{\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}}$$
Now I need to solve $\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}$, but I don't know how to go on.
 A: $$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}} = \lim_{x \to 1}e^ {{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}$$
Now we will check 
\begin{align*}
     \lim_{x \to 1} {{\tan\frac{\pi x}{2}}\ln\left(\tan\frac{\pi x}{4}\right)} 
    &\stackrel{\text{(L'Hôpital's rule)}}{=}
    \lim_{x \to 1} \frac{-\frac{\pi}{4}\tan^2(\frac{x\pi}{2})\cos^2(\frac{x\pi}{2})}{\cos^2(\frac{x\pi}{4})\tan(\frac{x\pi}{4})} \\&= \frac{\frac{-1}{2}}{\frac{1}{2}} = -1
\
\end{align*}
and therefore:
$\lim_{x \to 1} e^{{{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}} = e^{-1}$
A: $$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}=\lim_{x\rightarrow1}\left(1+\tan\frac{\pi x}{4}-1\right)^{\frac{1}{\tan\frac{\pi x}{4}-1}\cdot\tan\frac{\pi x}{2}\left(\tan\frac{\pi x}{4}-1\right)}=$$
$$=e^{-\lim\limits_{x\rightarrow}\frac{2\tan\frac{\pi x}{4}}{1+\tan\frac{\pi x}{4}}}=e^{-1}=\frac{1}{e}.$$
I used $\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=-\frac{2\tan\alpha}{(\tan\alpha-1)(1+\tan\alpha)}.$
A: Let $\tan\dfrac\pi4+1=u$ and using $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$
$$\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$$
$$=\lim_{u\to0}(1+u)^{\frac{2(u-1)}{1-(u-1)^2}}$$
$$=\left(\lim_{u\to0}(1+u)^{1/u}\right)^{\lim_{u\to0}\frac{2u(u-1)}{2u-u^2}}$$
Clearly, $\lim_{u\to0}(1+u)^{1/u}=e$ and 
$$\lim_{u\to0}\frac{2u(u-1)}{2u-u^2}=\lim_{u\to0}\frac{2(u-1)}{2-u}=?$$
A: Using your formula you have already covered the hard part. To proceed further just put $t=\tan (\pi x/4)$ so that $\tan(\pi x/2)=2t/(1-t^{2})$. The limit at the end of your question is $$\lim_{t\to 1}(t-1)\cdot\dfrac{2t}{1-t^{2}}=\lim_{t\to 1}-\frac{2t}{1+t}=-1$$ and the desired answer is $e^{-1}=1/e$.

By the way "not thinking about logics..." and "solving as many problems..." is not the way to beat competition because almost everyone is trying these easy approaches. You should go for "understanding basics" and this is the ability which separates the wheat from chaff in such competitions.
