$f(x)= \frac{\sin(\pi x)}{x(1-x)}$
How can I define $f(0)$ and $f(1)$ to make $f(x)$ continuous on $[0,1]$?
I've found that the limit at $0 = \pi$, and the limit from the left at $1 = \infty$.
I understand that if $f(0)=\pi$ then $f(x)$ is continuous on $[0,1)$, but must I define $f(1)=\infty$? Would that make $f(x)$ continuous?
EDIT: Also, can someone explain why $\lim\limits_{x \to 0} \frac{\sin\pi x}{x} = \pi$?