Prove continuouty of 1/x (delta-epsilon)

[wglmb]

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

Homework Statement

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

Homework Equations

The Attempt at a Solution

Just the first part for now - proving continuity on $\latexbb{R}\smallsetminus\{0\}$.
They want an $\epsilon-\delta$ proof. Here we go...


Let $a,b\in\latexbb{R}\smallsetminus\{0\}$.
Let $\delta=\epsilon>0$.
Then $\forall a\in\latexbb{R}\smallsetminus\{0\}$ and $\forall\epsilon>0$ we have:
$\mid a-b\mid<\delta\Rightarrow\mid g(a)-g(b)\mid=\mid \frac{1}{a}-\frac{1}{b}\mid=\mid\frac{b-a}{ab}\mid<\frac{\delta}{\mid ab\mid}$
Now I take the case where $b$ is such that $\mid a-b\mid>1$ and we have...
$\frac{\delta}{\mid ab\mid}<\delta=\epsilon$
So it's proven for such $b$ (I think?)

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

 

[Landau]

It is not a good idea to split cases. You will have to take a different delta to handle the situation at once. The thing that gives some trouble, is the denominator |ab|. We want to make sure this is small. Now, a is fixed (since we want to prove continuity at the fixed point a), so in fact the only trouble is 1/|b|, which we want to make small. This is the same as |b| being not too big. This can be arranged by choosing the right delta, which will depend on a. In that way, |a-b|<delta implies that b will be close to a, and after some fiddling you will be able to estimate 1/|b|.
I hope this helps you. You can also take a look at this thread.