Como solucionar limite trigonométrico que tiende a 1

Se me ocurre reemplazar la tangente por seno/coseno y luego intentar usar el limite conocido que

sen(x)/x = 1. Veamos...

$$\begin{align}&\lim_{x \to 1} \bigg((x-1) tan(\frac{\pi}2x) \bigg)=\\&\lim_{x \to 1} \bigg((x-1) \frac{\sin(\frac{\pi}2x)}{\cos(\frac{\pi}2x)} \bigg)\\&\text{Cambio variable: }\\&x-1=u \\&\lim_{u \to 0} \bigg(u \frac{\sin(\frac{\pi}2 (u+1))}{\cos(\frac{\pi}2 (u+1))} \bigg)=\\&\lim_{u \to 0} \bigg(u \frac{\sin(\frac{\pi u}2 + \frac{\pi}2)}{\cos(\frac{\pi u}2 + \frac{\pi}2)} \bigg)=\\&\lim_{u \to 0} \bigg(u \frac{\sin(\frac{\pi u}2)\cos( \frac{\pi}2)+ \cos(\frac{\pi u}{2})\sin(\frac{\pi}2)}{\cos(\frac{\pi u}2)\cos( \frac{\pi}2)-\sin(\frac{\pi u}2)\sin( \frac{\pi}2)} \bigg)=\\&\lim_{u \to 0} \bigg(u \frac{\cos(\frac{\pi u}{2})}{-\sin(\frac{\pi u}2))} \bigg)=\\&\text{multiplico y divido por }\pi/2\\&\lim_{u \to 0} \bigg(u \frac{\cos(\frac{\pi u}{2})}{-\sin(\frac{\pi u}2))} \cdot \frac{\frac{\pi}2}{\frac{\pi}2}\bigg)=\\&Reacomodo\\&\lim_{u \to 0} \bigg(-\frac{u \pi}{2} \cdot \frac{2}{\pi}\cdot \frac{\cos(\frac{\pi u}{2})}{\sin(\frac{\pi u}2))} \bigg)=\\&Reacomodo\\&\lim_{u \to 0} \bigg(- \frac{2}{\pi}\cdot \frac{\cos(\frac{\pi u}{2})}{\frac{\sin(\frac{\pi u}2))}{\frac{u \pi}{2}}} \bigg) \to\\&\to \bigg(- \frac{2}{\pi}\cdot \frac{1}{1} \bigg) \to-\frac{2}{\pi}\end{align}$$

Revisa los cálculos porque puedo haber cometido un error, pero creo que podrás tomar la idea

Salu2