How do I compute $\lim_{x \to a}(a-x)\tan\left(\frac{πx}{2a}\right)$

Question

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.

Answer 1

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: $$ \lim_{t\to 0}\frac{t}{\tan\left(\frac{\pi t}{2a}\right)} = \lim_{t\to 0}\frac{2ta}{\pi t} = \fbox{$\displaystyle \frac{2a}{\pi}$} $$

Answer 2

Let $a-x=y$

For $a\ne0,$

$$\tan\dfrac{\pi x}{2a}=\tan\dfrac{\pi(a-y)}{2a}=\cot\dfrac{\pi y}{2a}$$

Use $\lim_{h\to0}\dfrac{\sin h}h=1$

Answer 3

$a \not =0$.

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 $\lim_{z \rightarrow 0}\dfrac{\sin z}{z}=1$.

Answer 4

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: $$\lim_{x\to a} \color{blue}{\cos\left(\dfrac{π}{2}\left(1-\frac xa\right)\right)}=1;\\ \lim_{x\to 0} \color{red}{\frac{x}{\sin x}}=\lim_{x\to 0} \frac{\sin x}{x}=1.$$