Continuity of $f(x)$ involving infinity

Question

$f(x)= \frac{\sin(\pi x)}{x(1-x)}$

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 $\lim\limits_{x \to 0} \frac{\sin\pi x}{x} = \pi$?

Answer 1

We are interested in $\lim_{x \to 1}\frac{\sin \pi x}{1-x}$. If we let $y=1-x$, this becomes $$\lim_{y \to 0}\frac{\sin \pi (1-y)}{y}=\lim_{y \to 0}\frac{\sin (\pi -\pi y)}{y}=\lim_{y \to 0}\frac{\sin \pi y}{y}$$

Answer 2

$$ \lim_{x\to0} \frac{\sin(\pi x)}{x(1-x)} = \pi = \lim_{x\to1} \frac{\sin(\pi x)}{x(1-x)}. $$ Both can be readily shown by L'Hopital's rule.

Notice that the two limits have to be equal because of symmetry: If you let $u = 1-x$, then $x$ becomes $1-u$, and $\sin(\pi x)$ becomes $\sin(\pi(1-u))$, which is the same as $\sin(\pi u)$ by a trigonometric identity.

There are various ways to show that $\lim\limits_{x\to0} \dfrac{\sin x}{x} = 1$. Once you've done that, then you can write $$ \lim_{x\to0} \frac{\sin(\pi x)}{x} = \lim_{x\to0} \pi \frac{\sin(\pi x)}{\pi x} = \pi \lim_{x\to0} \frac{\sin(\pi x)}{\pi x} = \pi \lim_{w\to0} \frac{\sin(w)}{w}, $$ where $w= \pi x$. Notice that as $x\to 0$, $w$ also approaches $0$, thus justifying the part that says "$w\to0$".

Answer 3

To follow up on your edit:

We start with $\frac {\sin(\pi x)}{x}$ , then we do a change of variable, say, $u:= \pi x$, and we get:

$$\frac {\sin u}{u/\pi}=\frac {\pi\sin u}{u}=\pi \frac{\sin u}{u}$$