# Thread: Prove continuouty of 1/x (delta-epsilon)

1. ## Prove continuouty of 1/x (delta-epsilon)

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

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

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

Let $a,b\in\mathbb{R}\smallsetminus\{0\}$.
Let $\delta=\epsilon>0$.
Then $\forall a\in\mathbb{R}\smallsetminus\{0\}$ 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?

2. Originally Posted by 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!

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

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

Let $a,b\in\mathbb{R}\smallsetminus\{0\}$.
Let $\delta=\epsilon>0$.
Then $\forall a\in\mathbb{R}\smallsetminus\{0\}$ 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?
When $|x-x_0|<1$,

$-1 $\implies \frac{1}{x_0x}<\frac{1}{x_0^2-x_0}$

Thus, let $\delta = \min\{(x_0^2-x_0)\epsilon,1\}$. Therefore,

$\left|\frac{1}{x_0}-\frac{1}{x}\right| = \left|\frac{x-x_0}{x_0x}\right| < \left|\frac{\delta}{x_0x}\right|<\left|\frac{\delt a}{x_0^2-x_0}\right| = \left|\frac{(x_0^2-x_0)\epsilon}{x_0^2-x_0}\right| = \epsilon$

Thus, $\frac{1}{x}$ is continuous.

For the second part, all you have to prove is that $\lim_{x\to0^-}\frac{1}{x}\neq\lim_{x\to0^+}\frac{1}{x}$, which is obvious.

3. Stunning, thank you so much!

4. Originally Posted by redsoxfan325
When $|x-x_0|<1$,

$-1 $\implies \frac{1}{x_0x}<\frac{1}{x_0^2-x_0}$
Thus, let $\delta = \min\{(x_0^2-x_0)\epsilon,1\}$. Therefore,
$\left|\frac{1}{x_0}-\frac{1}{x}\right| = \left|\frac{x-x_0}{x_0x}\right| < \left|\frac{\delta}{x_0x}\right|<\left|\frac{\delt a}{x_0^2-x_0}\right| = \left|\frac{(x_0^2-x_0)\epsilon}{x_0^2-x_0}\right| = \epsilon$
There are several mistakes in the above.
It must be specified that $x_0>0$ because otherwise $-1 is false.

Here is another mistake: even if $x_0>0$ unless $x_0>1$ we do not know that $x_0^2-x_0>0$
so $\frac{1}{x_0x}<\frac{1}{x_0^2-x_0}$ may not be true.

To correct these mistakes you need two cases: $x_0\ge 1$ and $0.

5. Originally Posted by Plato
Here is another mistake: even if $x_0>0$ unless $x_0>1$ we do not know that $x_0^2-x_0>0$
so $\frac{1}{x_0x}<\frac{1}{x_0^2-x_0}$ may not be true.
OK, the first mistake was something that I forgot to mention. The proof is analogous if $x_0<0$.

And the second mistake I believe is taken care of by the absolute value signs in the actual delta-epsilon proof.

6. Originally Posted by redsoxfan325
And the second mistake I believe is taken care of by the absolute value signs in the actual delta-epsilon proof.
Actually no. That will not work.
Cases are necessary in this well worn proof.