Find the following limit $$L=\lim_{x\to 1}\left(2-x\right)^{\tan\frac{\pi x}{2}}.$$ Without L'Hopital's rule!
My try: $$L=\lim_{x\to1} e^{\ln \left[(2-x)^{\tan\frac{\pi x}{2}}\right]}=\lim_{x\to 1} e^{\tan\frac{\pi x}{2}\ln(2-x)}$$ Now we want to find $$\lim_{x\to1} \tan\frac{\pi x}{2}\ln(2-x)=\ln1\lim_{x\to1}\tan \dfrac{\pi x}{2}$$ I am stuck here.