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?

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- Nov 23rd 2006, 05:46 PMJacobpm64Epsilon-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 PMThePerfectHacker
- Nov 23rd 2006, 06:40 PMJacobpm64
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 PMThePerfectHacker
- Nov 23rd 2006, 06:54 PMJacobpm64
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 PMThePerfectHacker
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 PMJacobpm64
wow, major headache.. I understand all your examples

Thing is. The only thing that's killing me on mine is that there is no specific function... I'm just given f(x)

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.

*sigh*