Hi guys, I've been trying to do this for a while but I'm not really getting anywhere. Hints would be much appreciated!

Prove that the function $\displaystyle g(x)=1/x$ is continuous on $\displaystyle \mathbb{R}\smallsetminus\{0\}$, but cannot be defined at the origin $\displaystyle 0$ in such a way that the resulting function is continuous on $\displaystyle \mathbb{R}$.The problem

Just the first part for now - proving continuity on $\displaystyle \mathbb{R}\smallsetminus\{0\}$.My attempt

They want an $\displaystyle \epsilon-\delta$ proof. Here we go...

Let $\displaystyle a,b\in\mathbb{R}\smallsetminus\{0\}$.

Let $\displaystyle \delta=\epsilon>0$.

Then $\displaystyle \forall a\in\mathbb{R}\smallsetminus\{0\}$ and $\displaystyle \forall\epsilon>0$ we have:

$\displaystyle \mid a-b\mid<\delta\Rightarrow\mid g(a)-g(b)\mid=\mid \frac{1}{a}-\frac{1}{b}\mid=\mid\frac{b-a}{ab}\mid<\frac{\delta}{\mid ab\mid}$

Now I take the case where $\displaystyle b$ is such that $\displaystyle \mid a-b\mid>1$ and we have...

$\displaystyle \frac{\delta}{\mid ab\mid}<\delta=\epsilon$

So it's proven for such $\displaystyle b$ (I think?)

Now I have no idea what to do about $\displaystyle b$ such that $\displaystyle \mid a-b\mid<1$. Or was splitting it into two cases a bad idea? Am I going anywhere useful here?

Please help!