$\displaystyle

f(x) = \left\{

\begin{array}{lr}

x^2 sin\frac{1}{x} & : x \neq 0\\

0 & : x = 0

\end{array}

\right.

$

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- Oct 22nd 2009, 10:27 PMbinkypoohow can I prove if this function is continuous or not!
$\displaystyle

f(x) = \left\{

\begin{array}{lr}

x^2 sin\frac{1}{x} & : x \neq 0\\

0 & : x = 0

\end{array}

\right.

$ - Oct 22nd 2009, 10:34 PMredsoxfan325
- Oct 25th 2009, 08:05 PMbinkypoo
doesnt this show that f is continuous at x = 0? What about the rest of the domain?

- Oct 25th 2009, 08:19 PMredsoxfan325
It's continuous everywhere else because $\displaystyle x^2\sin(x)$ and $\displaystyle \frac{1}{x}$ are. The composition of continuous functions is continuous.

- Oct 25th 2009, 08:33 PMbinkypoo
I guess my problem is I dont know how to show that $\displaystyle x^2sin(x)$ and 1/x are continuous EVERYWHERE. Could you explain that please? thanks

- Oct 25th 2009, 08:54 PMredsoxfan325
I don't think your professor expects you to do all that, but if he does, you need to crank out the $\displaystyle \epsilon$-$\displaystyle \delta$ proofs for each one.

For $\displaystyle x^2$ let $\displaystyle \delta=\min\left\{1,\frac{1}{1+2|x_0|}\right\}$, because $\displaystyle |x-x_0|<1\implies |x+x_0|<1+2|x_0|\implies |x-x_0||x+x_0|<\delta(1+2|x_0|)$

For $\displaystyle \sin x$, you can probably Google a proof. The idea is to let $\displaystyle \delta=\epsilon$ and then use a difference-to-product identity to convert $\displaystyle |\sin x-\sin x_0|$ into something more manageable.

$\displaystyle \frac{1}{x}$ is kind of a pain in the ass, and I've screwed up continuity proofs of this function before. There are probably other threads on MHF with proofs of this.