Find the limit of the complex function. $$ \lim_{x\to a} \left(2- \frac{x}{a}\right)^{\left(\tan \frac{\pi x}{2a}\right)}.$$
I have simplified this limit to this extent :
$$e^{ \lim_{x\to a} \left(\left(1- \frac{x}{a}\right){\left(\tan \frac{\pi x}{2a}\right)}\right)}$$
I don't know how to simplify the limit after that. please help.
 A: \begin{align*}
\log\left(2-\frac{x}{a}\right)^{\tan(\pi x/2a)}&=\left(\tan\frac{\pi x}{2a}\right)\left(\log\left(2-\frac{x}{a}\right)\right)\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{\log\left(1+\left(1-\dfrac{x}{a}\right)\right)}{\cos\left(\dfrac{\pi}{2}\left(\dfrac{x}{a}-1\right)+\dfrac{\pi}{2}\right)}\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{\left(1-\dfrac{x}{a}\right)-\dfrac{1}{2}\left(1-\dfrac{x}{a}\right)^{2}+\cdots}{\dfrac{\pi}{2}\left(1-\dfrac{x}{a}\right)-\dfrac{1}{3!}\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a}\right)\right)^{3}+\cdots}\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{1-\dfrac{1}{2}\left(1-\dfrac{x}{a}\right)+\cdots}{\dfrac{\pi}{2}-\dfrac{1}{3!}\left(\dfrac{\pi}{2}\right)^{3}\left(1-\dfrac{x}{a}\right)^{2}+\cdots},
\end{align*}
taking limit as $x\rightarrow a$, then $\log\left(2-\dfrac{x}{a}\right)^{\tan(\pi x/2a)}\rightarrow\dfrac{2}{\pi}$.
A: $$ \lim_{x\to a} \left(2- \frac{x}{a}\right)^{\tan \frac{\pi x}{2a}}= \lim_{x\to a} \left(1+1- \frac{x}{a}\right)^{\frac{1}{1-\frac{x}{a}}\cdot\left(1-\frac{x}{a}\right)\tan \frac{\pi x-\pi a+\pi a}{2a}}=$$
$$=e^{\frac{2}{\pi}\lim\limits_{x\rightarrow a}\left(\frac{\frac{\pi}{2a}(a-x)}{\sin\frac{\pi}{2a}(a-x)}\cdot\cos\frac{\pi}{2a}(a-x)\right)}=e^{\frac{2}{\pi}}.$$
A: Suppose $y \triangleq \frac{\pi x}{2a}$. Then, we have
$$\lim_{t \rightarrow \pi/2} \tan(y) (1-\frac{2y}{\pi})$$
Then, using the Taylor series expansion and simple manipulations we have:
$$\lim_{t \rightarrow \pi/2} \frac{1}{\pi} \Big(-\frac{1}{y-\pi/2}+\frac{1}{3}(y-\pi/2) + \ldots \Big)(\pi - 2y)= \frac{2}{\pi}.$$
Then: 
$$ \lim_{x\to a} \left(2- \frac{x}{a}\right)^{\left(\tan \frac{\pi x}{2a}\right)} = e^{\frac{2}{\pi}}$$
A: From the given limit, using $t = x-a$, you can rewrite it as:
$$\lim_{t\to 0}\left(1-\frac{t}{a}\right)^{-\cot\left(\frac{\pi t}{2a}\right)}$$
This can be converted to the form $e^{\dots}$ using the definition of exponent function, by multiplying and dividing in the power by $-t/a$. Thus we have:
$$\lim_{t\to 0}\left(1-\frac{t}{a}\right)^{-\cot\left(\frac{\pi t}{2a}\right) \times \frac{-a}{t} \times \frac{-t}{a}} =\lim_{t \to 0} e^{\cot\left(\frac{\pi t}{2a}\right) \times \frac{t}{a}}$$
Then due continuity of $e^x$, we have:
$$\begin{align}\lim_{t \to 0} e^{\cot\left(\frac{\pi t}{2a}\right) \times \frac{t}{a}} &=  e^{\lim_{t \to 0}\cot\left(\frac{\pi t}{2a}\right) \times \frac{t}{a}} \\
&= \exp\left(\lim_{t \to 0}\frac{\frac{t}{a} \times\frac{\pi}{2} \times \frac{2}{\pi}}{\tan\left(\frac{\pi t}{2a}\right)}\right) \\
&= \exp\left(\frac{2}{\pi}\right)
\end{align}$$
A: Let $y=(2- \dfrac xa)^{\tan \frac{\pi x}{2a}}$ then $\ln y=\tan\dfrac{\pi x}{2a}\ln(2- \dfrac xa)$ and let $1-\dfrac{x}{a}=\dfrac{1}{t}$ then
\begin{align}
\lim_{x\to a} (2- \dfrac xa)^{\tan \frac{\pi x}{2a}}
&= \lim_{t\to \infty} \tan\dfrac{\pi}{2}\left(1-\dfrac1t\right)\ln\left(1+\dfrac1t\right) \\
&= \lim_{t\to \infty} \cot\dfrac{\pi}{2t}\ln\left(1+\dfrac1t\right) \\
&= \lim_{t\to \infty} \dfrac{\cos\dfrac{\pi}{2t}}{\sin\dfrac{\pi}{2t}}\ln\left(1+\dfrac1t\right) \\
&= \lim_{t\to \infty} \dfrac{\dfrac{\pi}{2t}\cos\dfrac{\pi}{2t}}{\sin\dfrac{\pi}{2t}} \cdot t\ln\left(1+\dfrac1t\right) \dfrac{1}{t\dfrac{\pi}{2t}}\\
&= \lim_{t\to \infty} \dfrac{\dfrac{\pi}{2t}}{\sin\dfrac{\pi}{2t}} \cdot \ln\left(1+\dfrac1t\right)^t ~~\dfrac{\cos\dfrac{\pi}{2t}}{\dfrac{\pi}{2}}\\
&= \color{blue}{\dfrac{2}{\pi}}
\end{align}
using $\lim_{t\to \infty}\left(1+\dfrac1t\right)^t=e$, then $y=e^{\frac{2}{\pi}}$.
A: Use a change of variable
write $$ \lim_{x\to a}(1 - x/a)\tan(\frac{\pi x}{ 2a}) = \lim_{x\to a} \frac{(1 - x/a)}{\cos(\frac{\pi x}{ 2a})}\sin(\frac{\pi x}{ 2a}) $$
To solve the indetermination make the change of variable: $ t = 1- \frac{x}{a}$. You have that $t \to 0$ when $x \to a$.
$$\lim_{t \to 0} \frac{t}{ \cos (\frac{\pi}{2}(1-t))} = \lim_{t \to 0} \frac{t}{ \cos(\frac{\pi}{2})\cos(\frac{\pi}{2}t) + \sin(\frac{\pi}{2})\sin (\frac{\pi}{2}t)} =  \lim_{t \to 0} \frac{t}{\sin (\frac{\pi}{2}t)}  = \frac{2}{\pi}$$
Por the last equation use that
$$\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$
$$ \frac{\sin(\nabla)}{\nabla} \to 1 \quad \text{ if } \quad \nabla \to 0 $$
