Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
Did you calculate the limit of the function as $x \to 1$? You have noticed that the denominator approaches $0$ as $x \to 1$. What about the numerator? – Srivatsan Sep 11 '11 at 16:34
The limit at $1$ is not $\infty$. – Chris Eagle Sep 11 '11 at 16:35
I think I'm having some trouble with the rules of sin limits. I see now that the limit at the top approaches 0 as well, which would make the limit 1? What I did to find the limit at 1 (incorrectly) was I separated the function and found the limits as x->1 of $\frac{sin(x)}{x}$ and $\frac{\pi}{x-1}$, and multiplied the two. If I did this incorrectly, does that mean the limit as x->0 of $\frac{sin(\pi x)}{x}$ doesn't equal $\pi$? – BKaylor Sep 11 '11 at 16:43
You cannot separate the x that way, since x is part of the argument (angle) in $sin( \ pi x)$. Continuity means that limits from the left and right at a point p are equal to each other, and both are equal to the value of the function at p. Have you worked with the limit of $\frac {sinx}{x}$ as $x \rightarrow 0$? – gary Sep 11 '11 at 16:51
Yes, I know that the limit as $x→0 \frac{sinx}{x} = 1$. Since my separation was incorrect, how can I show that the lim as x->0 of $\frac{sin(\pi x)}{x(1-x)} = \pi$? – BKaylor Sep 11 '11 at 16:57
up vote 3 down vote accepted

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

share|cite|improve this answer
I would consider $\sin(\pi-x)=\sin(x)$ to be more basic than using the addition formula, because it amounts to reflecting across the vertical axis. Alternatively, combining $\sin(\pi+x)=-\sin(x)$ and $\sin(-x)=-\sin(x)$ would work, and these can be more readily seen than the addition formula. That's nothing against this answer; the nice thing about the addition formula is that it can be specialized to prove most other identities of $\sin$. – Jonas Meyer Sep 11 '11 at 17:30
I don't know if you cancelled somehow, but the denominator is actually x(1-x). – BKaylor Sep 11 '11 at 17:49
@BKaylor: But near x=1 that term doesn't cause a problem. I just focused on the terms that do. – Ross Millikan Sep 11 '11 at 18:48
@Jonas Meyer: Good point. I think that is better. This is what came to mind at the time. – Ross Millikan Sep 11 '11 at 18:49
@Ross Millikan: I understand, since it approaches 1. Thanks for your help! – BKaylor Sep 11 '11 at 18:52

$$ \lim_{x\to0} \frac{\sin(\pi x)}{x(1-x)} = \pi = \lim_{x\to1} \frac{\sin(\pi x)}{x(1-x)}. $$ 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 "$w\to0$".

share|cite|improve this answer
Thanks for your answer!! Can you expand on the limit as x ->1 in the first equation? How can you algebraically show that it $=\pi$? – BKaylor Sep 11 '11 at 17:55
Also, because of this, the Intermediate Value Theorem cannot be applied since f(a)=f(b) on [a,b], correct? – BKaylor Sep 11 '11 at 18:23

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:

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.