Evaluate without L'Hopital: $\lim_{x \to 3} (x-3)\csc\pi x $

Question

I'm supposed to evaluate this limit without using L'Hopital's rule.

$$\lim_{x \to 3} (x-3)\csc\pi x $$

I find the indeterminate form of $0$ or $\frac{0}{0}$. The latter tells me that L'Hopital's is an option, but since we haven't seen derivatives yet I'm not allowed to used it.

Previously I already tried swapping the $\csc\pi x$ for $\frac{1}{\sin\pi x}$ but when doing this I can't seem to get rid of the sinus. I also believe that since the limit goes to $3$ and not to $0$, the $\lim_{x \to 0}\frac{\sin ax}{ax} = 1$ rule is not an option either.

I tried making the $\sin(\frac{\pi}{2})$ so that $=1$, but without any success. Can anyone give me a hint? I don't need a full solution as I want to try and find it myself.

Answer 1

(Edited because I just realised you don't want a full solution)

Hint: $\sin(x) = -\sin(x + 3\pi)$. Try a substitution such that the limit you mentioned is an option!

Answer 2

By a change of variable,

$$\lim_{x\to3}\frac{x-3}{\sin\pi x}=\lim_{y\to0}\frac{y}{\sin(\pi y+3\pi)}=-\frac1\pi\lim_{y\to0}\frac{\pi y}{\sin(\pi y)}.$$