Evaluate the limit $\lim_{x\to a} (a-x) \tan \frac {\pi x}{2a}$

Question

Evaluate: $$\lim_{x\to a} (a-x) \tan \dfrac {\pi x}{2a}.$$

My attempts: $$=\lim_{x\to a} (a-x) \dfrac {\sin \left(\dfrac {\pi x}{2a}\right)}{\cos \left(\dfrac {\pi x}{2a}\right)}=1\cdot\lim_{x\to a} \dfrac {a-x}{\cos \left(\dfrac {\pi x}{2a}\right)}.$$

Answer 1

Note that, by letting $t=a-x$, $$\lim_{x\to a} (a-x) \tan \dfrac {\pi x}{2a}= \lim_{t\to 0} t \tan \dfrac {\pi a-\pi t}{2a}= \lim_{t\to 0} t \tan \left(\dfrac {\pi}{2} -\frac{\pi t}{2a}\right)= \lim_{t\to 0} \frac{t}{\tan \left(\frac{\pi t}{2a}\right)}=\frac{2a}{\pi}$$ where in the last step we used the fact that $\tan(x)/x\to 1$ as $x\to 0$.

Answer 2

$$\lim_{x\to a}(a-x)\tan\frac{\pi x}{2a} = \lim_{x\to a}(a-x)\frac{\sin\frac{\pi x}{2a}}{\cos \frac{\pi x}{2a}} = \lim_{x\to a}(a-x)\frac{\sin\frac{\pi x}{2a}}{\sin (-\frac{\pi x}{2a}+\frac{\pi}{2})} = \frac{2a}{\pi}\lim_{x\to a}\frac{\frac{\pi}{2a}(a-x)}{\sin(\frac{\pi}{2a}(a-x))} = \frac{2a}{\pi}.$$

Answer 3

Let $a-x=y$.

Thus, $x=a-y$ and $$\lim_{x\to a} (a-x) \tan \dfrac {\pi x}{2a}=\lim_{y\rightarrow0}y\cot\frac{\pi y}{2a}=\frac{2a}{\pi}\lim_{y\rightarrow0}\frac{\frac{\pi y}{2a}}{\sin\frac{\pi y}{2a}}=\frac{2a}{\pi}$$

Answer 4

Let $y = \frac{\pi x}{2a}$

\begin{align} \lim_{x \to a} \frac{a-x}{\cos \left( \frac{\pi x}{2a}\right)} &= \lim_{y \to \frac{\pi}{2}} \frac{\pi a-(2ay)}{\pi \cos(y)}\\ &=\frac{2a}{\pi} \lim_{y \to \frac{\pi}{2}} \frac{\frac{\pi}2-y}{\cos(y)}\\ &=\frac{2a}{\pi} \lim_{y \to \frac{\pi}{2}} \frac{\frac{\pi}2-y}{\cos\left(y-\frac{\pi}{2} + \frac{\pi}{2}\right)}\\ &= \frac{2a}{\pi} \lim_{y \to \frac{\pi}{2}} \frac{y-\frac{\pi}2}{\sin\left(y-\frac{\pi}{2} \right)}\\ &=\frac{2a}{\pi}\end{align}