# Epsilon-Delta

• Nov 23rd 2006, 05:46 PM
Jacobpm64
Epsilon-Delta
Express the following as an Epsilon-Delta proof (to show that it is continuous):

$\displaystyle \lim_{x\rightarrow - \infty}f(x) = L$

Can someone give me ideas for this one?
• Nov 23rd 2006, 06:08 PM
ThePerfectHacker
Quote:

Originally Posted by Jacobpm64
Express the following as an Epsilon-Delta proof (to show that it is continuous):

$\displaystyle \lim_{x\rightarrow - \infty}f(x) = L$

Can someone give me ideas for this one?

Continous has nothing to do with that.

But it means.
For any $\displaystyle \epsilon >0$
There exists a $\displaystyle N$.
If,
$\displaystyle x<N$ is in the domain
Then,
$\displaystyle |f(x)-L|<\epsilon$
• Nov 23rd 2006, 06:40 PM
Jacobpm64
all right, I understand where you got the x < N, because x doesn't approach a specific value, but it does have to be less than some number N.

So, usually I'd write
$\displaystyle |f(x) - L| < \epsilon \quad \Rightarrow \quad |f(x) - L| < \epsilon$

Then I'd try to get that in the form $\displaystyle |x - c| < \delta$

How do I do that here?
• Nov 23rd 2006, 06:50 PM
ThePerfectHacker
Quote:

Originally Posted by Jacobpm64
all right, I understand where you got the x < N, because x doesn't approach a specific value, but it does have to be less than some number N.

So, usually I'd write
$\displaystyle |f(x) - L| < \epsilon \quad \Rightarrow \quad |f(x) - L| < \epsilon$

Then I'd try to get that in the form $\displaystyle |x - c| < \delta$

How do I do that here?

There is no delta here. It is a different type of limit.
Maybe you mean this.
$\displaystyle \forall \epsilon >0, \exists N \, \, \, \, \, x<N\to |f(x)-L|<\epsilon$
• Nov 23rd 2006, 06:54 PM
Jacobpm64
Oh, all right.. No delta because x is extending without bound towards negative infinity, but there is an epsilon because the function approaches a finite value of L, correct?

I understand that.

But, I still don't know how I'd go about constructing a proof.
• Nov 23rd 2006, 07:05 PM
ThePerfectHacker
Quote:

Originally Posted by Jacobpm64

But, I still don't know how I'd go about constructing a proof.

Say you want to show,
$\displaystyle \lim_{x\to -\infty} \frac{1}{x}=0$

You need to show that for any $\displaystyle \epsilon>0$
We can choose an $\displaystyle N$
Such that if,
$\displaystyle x<N$ is in the domain of the function.
Implies,
$\displaystyle \left|\frac{1}{x}\right|<\epsilon$
Since both side are positive we can take the reciprocal of both sides, but then we have to flip the inequality, obtaining an equivalent statement,
$\displaystyle |x|<\frac{1}{\epsilon}$
That means, $\displaystyle x<\frac{1}{\epsilon}$.
We also note that,
$\displaystyle \frac{1}{\epsilon}$ is in the domain of $\displaystyle f(x)$.

So this tell us given any $\displaystyle \epsilon$ we need to choose $\displaystyle N=\frac{1}{\epsilon}$
• Nov 23rd 2006, 07:24 PM
Jacobpm64
In your example, you wrote $\displaystyle |f(x) - L| < \epsilon$
as $\displaystyle x < \frac{1}{ \epsilon }$
That isn't too bad to do because f(x) is defined as a specific function, $\displaystyle \frac{1}{x}$
I don't know how I'd write $\displaystyle |f(x) - L| < \epsilon$ in the form x < something, when I don't have any specific function, so I only have an f(x), not an x.