Prove that $f(x)=x^2 \cdot \lfloor {\frac{1}{x^2}}\rfloor$ is continuous.

Question

I want to prove that $f(x)=x^2 \cdot \lfloor {\frac{1}{x^2}}\rfloor$ is continuous on its domain.

This is what I have come up with so far : I know that the floor function is continuous over $\mathbb{R}-\mathbb{Z}$,and I think this implies that $f$ is continuous over $\mathbb{R}-\mathbb{Z}$. I think that $f(x) =0$,$\forall x \in \mathbb{Z} $ and as a result $f$ is continuous on its domain. Is this right?

Answer 1

I think your function is not continuous, unfortunately.

For instance, $f(1)= 1$, but for all $x > 1$, you have $f(x) = 0$.

Answer 2

Is it continuosus? This is the graph:

It is not continuous at $x=1$ which is in domain.

$$\lim_{x\to 1_{-} } f(x) =1 \ne 0 =\lim_{x\to 1_{+} } f(x)$$