I need help solving a trigonometric limit.

Question

I'm seriously stuck. The limit is $(x-3)\,\csc(\pi x)$ where $x$ approaches $3$.

Also, please don't make use of l'Hôpital's rule.

Thanks in advance.

Answer 1

Note that $\sin(\pi x)=\sin(\pi(x-3+3))=-\sin(\pi(x-3))$. Then we can write

$$\frac{x-3}{\sin(\pi x)}=-\frac{x-3}{\sin(\pi(x-3))}$$

Letting $y=\pi(x-3)$ we have

$$\lim_{x\to 3}\frac{x-3}{\sin(\pi x)}=-\frac{1}{\pi}\lim_{y\to 0}\frac{y}{\sin(y)}=-\frac1\pi$$