# Finding points at which $f(x)=\frac{{{\sin }^{2}}x}{x|x(\pi -x)|}$ is not continuous

I have a homework question to find the points at which $$f(x)=\frac{{{\sin }^{2}}x}{x|x(\pi -x)|}$$ is not continuous. I can't seem to figure it out for this function.

Can someone help me find these points?

Thanks!

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What exactly is your difficulty in finding the points of discontinuity? What have you tried? What do you know about the points of discontinuity for a function of the form $f(x)/g(x)$? –  JavaMan Nov 25 '11 at 20:43
Well $g(x)$ cant be $0$, so are the only points the points $x=0$ and $x=\pi$? –  Jason Nov 25 '11 at 20:47
I would suggest to first find the places of discontinuity. Since your function is of the form $f(x) = h(x)/g(x)$, where $g$ and $h$ are both continuous, you should have a pretty good idea about where $f$ is discontinuous. Then, by whatever means you are required, you probably should show that if you stay away from the discontinuities, then your function is continuous everywhere else. –  JavaMan Nov 25 '11 at 20:53

A function $f(x)$ is continuous at $x=a$ if and only if it satisfies three conditions:

1. $f$ is defined at $x=a$;
2. $\lim\limits_{x\to a}f(x)$ exists; and
3. $\lim\limits_{x\to a}f(x) = f(a)$.

By the usual limit laws and the definitions of sums, products, quotients, and compositions of functions, we have:

1. If $f$ and $g$ are each continuous at $a$, then so are $f+g$, $f-g$, and $fg$.
2. If $f$ and $g$ are each continuous at $a$, and $g(a)\neq 0$, then $\frac{f}{g}$ is continuous at $a$.
3. If $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g\circ f$ is continuous at $a$.

Now, $f(x)=\sin(x)$ is continuous everywhere, so the product of $f$ with itself, $\sin^2(x)$, is continuous everywhere.

The constant function $f(x)=\pi$ is continuous everywhere, the identity function $g(x)=x$ is continuous everywhere, so their difference, $x\mapsto \pi - x$ is continuous everywhere.

The function $f(x)=\pi -x$ is continuous everywhere, the identity function $g(x)=x$ is continuous everywhere, so their product $fg(x) = x(\pi -x)$ is continuous everywhere.

The function $f(x)=x(\pi-x)$ is continuous everwhere, the function $g(x)=|x|$ is continuous everywhere, so their composition $g\circ f(x) = |x(\pi-x)|$ is continuous everywhere.

Similarly, the function $x\mapsto x|x(\pi - x)|$ is continuous eveywhere.

So: the numerator of your function is continuous everywhere. The denominator of your function is continuous everywhere. So... where is their quotient continuous?

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At all the points where $x|x(\pi-x)|$ is not $0$ i.e the function is defined :) –  Jason Nov 25 '11 at 21:29