# Thread: how can I prove if this function is continuous or not!

1. ## how can I prove if this function is continuous or not!

$\displaystyle f(x) = \left\{ \begin{array}{lr} x^2 sin\frac{1}{x} & : x \neq 0\\ 0 & : x = 0 \end{array} \right.$

2. Originally Posted by binkypoo
$\displaystyle f(x) = \left\{ \begin{array}{lr} x^2 sin\frac{1}{x} & : x \neq 0\\ 0 & : x = 0 \end{array} \right.$
Take the limit: $\displaystyle -\lim_{x\to0}x^2\leq\lim_{x\to0}x^2\sin(1/x)\leq\lim_{x\to0}x^2$

So $\displaystyle \lim_{x\to0}x^2\sin(1/x)=0=f(0)$, and it's continuous at $\displaystyle x=0$.

It's also differentiable at $\displaystyle x=0$.

3. doesnt this show that f is continuous at x = 0? What about the rest of the domain?

4. It's continuous everywhere else because $\displaystyle x^2\sin(x)$ and $\displaystyle \frac{1}{x}$ are. The composition of continuous functions is continuous.

5. 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

6. Originally Posted by binkypoo
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
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.