How do I evaluate $\lim\limits_{z \to 1}\left(1-z\right)\tan{\dfrac{\pi z}{2}}$?

Question

The question is from Differential and Integral Calculus by Piskunov:

$$\lim\limits_{z \to 1}\left(1-z\right)\tan{\dfrac{\pi z}{2}}$$

I tried using the identity, $\tan{\frac{x}{2}}=\frac{1-\cos{x}}{\sin{x}}$ to simplify this to:

$$\lim\limits_{z \to 1}\left(1 - z\right)\dfrac{\sin{\pi z}}{1+\cos{\pi z}}$$

But I still can't solve this.

Answer 1

Note that\begin{align}\tan\left(\frac{\pi z}2\right)&=\tan\left(\frac{\pi(z-1)}2+\frac\pi2\right)\\&=-\cot\left(\frac\pi2(z-1)\right),\end{align}and that therefore\begin{align}\lim_{z\to1}(1-z)\tan\left(\frac{\pi z}2\right)&=\lim_{z\to1}\left(\frac{z-1}{\sin\left(\frac\pi2(z-1)\right)}\times\cos\left(\frac\pi2(z-1)\right)\right)\\&=\frac{\lim_{z\to1}\cos\left(\frac\pi2(z-1)\right)}{\lim_{z\to1}\frac{\sin\left(\frac\pi2(z-1)\right)}{z-1}}\text{ (since both limits exist)}\\&=\frac{\cos0}{\frac\pi2\cos0}\\&=\frac2\pi.\end{align}

Answer 2

$$\lim_{z \to 1}\frac{\cos(\frac{\pi z}{2})}{1-z} = -\lim_{z \to 1}\frac{\cos(\frac{\pi z}{2}) -\cos(\frac{\pi}{2})}{z-1} = -\cos(\frac{\pi x}{2})' \left|\frac{}{}_{x=1} \right. =\frac{\pi}{2}\sin(\frac{\pi}{2}) = \frac{\pi}{2}$$ and $$\lim_{z \to 1}\sin(\frac{\pi z}{2}) = 1 $$ Combining the above two $$\lim_{z \to 1}(1-z)\tan(\frac{\pi z}{2}) = \lim_{z\to 1}\frac{1-z}{\cos(\frac{\pi z}{2})} \cdot \lim_{z \to 1}\sin(\frac{\pi z}{2}) = \frac{2}{\pi}\cdot 1$$

Answer 3

Let $t:=\dfrac\pi2(1-z)$.

$$\lim\limits_{z \to 1}\left(1-z\right)\tan{\dfrac{\pi z}{2}}=\frac2\pi\lim\limits_{t \to 0}t\cot t=\frac2\pi\lim\limits_{t \to 0}\frac t{\sin t}\cdot\lim\limits_{t \to 0}\cos t.$$