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

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

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2  
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
2  
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

3 Answers 3

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

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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

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:

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

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

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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

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