# 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.

• Do you know that you can "accept" one of the given answers? Please take a few minutes for a tour: math.stackexchange.com/tour – Robert Z Dec 12 '18 at 16:25

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

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

Let $$a-x=y$$

For $$a\ne0,$$

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

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

$$a \not =0$$.

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