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

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

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

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

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

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 $x^2\sin(x)$ and $\frac{1}{x}$ are. The composition of continuous functions is continuous.

5. I guess my problem is I dont know how to show that $x^2sin(x)$ 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 $x^2sin(x)$ 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 $\epsilon$- $\delta$ proofs for each one.

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

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

$\frac{1}{x}$ 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.