Finding $ \lim_{x \to 1} \frac{\tan \pi x}{1-x} $ without L’Hôpital

Question

How to calculate $$ \lim_{x \to 1} \frac{\tan \pi x}{1-x} $$ WITHOUT using L'Hôpital's rule?

Answer 1

Let $x=1+y.$ If $0<|y|<1/2$ then $$\frac {\tan \pi x}{1-x}=\frac {1}{1-\tan \pi y}\cdot\frac {\tan \pi y}{\pi y}\cdot \pi$$ and $y\to 0$ as $x\to 1.$