How to compute $\lim_{x \to 1} \frac{\sin^a{\pi x}}{(1-x)^a} , a \in \Re$

Question

I'm trying to compute this limit:

$$ \lim_{x \to 1} \frac{\sin^a{\pi x}}{(1-x)^a} , a \in \Re $$

It should be quite easy, however, I just can't see it now... if it would be without $a$, it would be very easy with using l'Hospital rule. However, it isn't possible to do in this situation.

I think that it would help if I expressed $\sin{\pi x}$ with using the cosine function somehow. However, as I've said, I can't see it, and maybe this is not a good idea...

Thank you for any help!

Answer 1

Let $y=1-x$. We then have $$\dfrac{\sin^a(\pi x)}{(1-x)^a} = \dfrac{\sin^a(\pi -\pi y)}{y^a}=\dfrac{\sin^a(\pi y)}{y^a}$$ Hence, the limit as $x \to 1$, is same as $y \to 0$, which is nothing but $\pi^a$.

Answer 2

Hints: $\sin \pi x $ ~ $\pi x$ when $x\to 1$