How do I compute $\lim_{x \to a}(a-x)\tan\left(\frac{πx}{2a}\right)$ Evaluate
$$\lim_{x \to a}f(x)=(a-x)\tan\left(\frac{πx}{2a}\right)$$
I tried changing the tangent into cotangent by writing it in the form of $\cot\left(\dfrac{π}{2}-\dfrac{πx}{2a}\right)$. Then I factored out $\dfrac{π}{2}$. But this didn't lead me to anything rigid. If anyone in the community could help me out, I'd really appreciate it.
 A: It does lead to a solution, keep going:
$$\begin{align}\lim_{x \to a}(a-x)\tan\left(\dfrac{πx}{2a}\right)&=\lim_{x \to a}(a-x)\cot\left(\dfrac{π}{2}\left(1-\frac xa\right)\right)=\\
&=\lim_{x \to a}(a-x)\cdot \frac{\color{blue}{\cos\left(\dfrac{π}{2}\left(1-\frac xa\right)\right)}}{\sin\left(\dfrac{π}{2}\left(1-\frac xa\right)\right)}=\\
&=\lim_{x \to a} \color{red}{\frac{\dfrac{π}{2}\left(1-\frac xa\right)}{\sin\left(\dfrac{π}{2}\left(1-\frac xa\right)\right)}}\cdot \frac{2a}{\pi}=\\
&=\frac{2a}{\pi},\end{align}$$
where it was used:
$$\lim_{x\to a} \color{blue}{\cos\left(\dfrac{π}{2}\left(1-\frac xa\right)\right)}=1;\\
\lim_{x\to 0} \color{red}{\frac{x}{\sin x}}=\lim_{x\to 0} \frac{\sin x}{x}=1.$$
A: Let $a-x=y$
For $a\ne0,$
$$\tan\dfrac{\pi x}{2a}=\tan\dfrac{\pi(a-y)}{2a}=\cot\dfrac{\pi y}{2a}$$
Use $\lim_{h\to0}\dfrac{\sin h}h=1$
A: $a \not =0$.
Set $y =(πx)/(2a).$
Then $y \rightarrow π/2.$
$((2a)/π)(π/2-y)\dfrac{\sin y}{\cos y}.$
Recall $\sin (π/2-y) =\cos y.$
Then 
$(2a/π)(\dfrac {1}{\dfrac{\sin (π/2-y)}{π/2-y}})(\sin y)$.
Can you take the limit $y \rightarrow π/2?$
P.S. Recall $\lim_{z \rightarrow 0}\dfrac{\sin z}{z}=1$.
A: As an alternative:
$$
\lim_{x \to a}f(x)=\lim_{x\to a}\left((a-x)\tan\left(\frac{πx}{2a}\right)\right)
$$
Substitute $t = x - a \iff x = t + a$, so your limit becomes:
$$
\lim_{t \to 0}f(t)=\lim_{t\to 0}\left((a-(t+a))\tan\left(\frac{π(t+a)}{2a}\right)\right) = \\
\lim_{t\to 0}\left((-t)\tan\left(\frac{\pi t + \pi a}{2a}\right)\right) = -\lim_{t\to 0}\left(t\tan\left(\frac{\pi t}{2a} + \frac{\pi}{2}\right)\right) = \\
= \lim_{t\to 0}\left(t\cot\left(\frac{\pi t}{2a}\right)\right) = \lim_{t\to0}\frac{t}{\tan\left(\frac{\pi t}{2a}\right)}
$$
Now by Taylor of $\tan x$ as $x\to 0$:
$$
\tan x \sim x \\
$$
We get:
$$
\lim_{t\to 0}\frac{t}{\tan\left(\frac{\pi t}{2a}\right)} = \lim_{t\to 0}\frac{2ta}{\pi t} = \fbox{$\displaystyle \frac{2a}{\pi}$}
$$
