Amy Sherfield

Posted on Monday, 12 April, 2004 - 01:08 pm:

I am a bit confused.

I have read the definition of a function continuous at a point $a$ as:

''For each $ε > 0$ , there exists a $δ > 0$ such that $| x - a | < δ \Rightarrow | f (x) - f (a) | < ε$ .''

If I had to prove that a function is continuous at a particular point, e.g. $x^{2}$ at 0, for example, I know that $δ$ , $ε > 0$ but can $ε > δ$ or $ε = δ$ or $ε < δ$ .

Which of these possibilities can occur for $ε$ and $δ$ from the definition I have written in quotes?

Thanx

AMY

Phil Freeman

Posted on Monday, 12 April, 2004 - 07:09 pm:

The stuff in quotes basically means tell me how close you want to be to

f (x)

and I'll tell you how close you have to be to

x

e.g. for $f (x) = x^{2}$ at $x = 0$ , choose $δ = \sqrt{ε}$ .

Then if

$| 0 - a | < δ$

$a < δ$

$a^{2} < ε$

$| 0^{2} - a^{2} | < ε$

$| f (0) - f (a) | < ε$

therefore continuous at x=0

-Phil.

pps. dear uni guys: never done that before, probably very wrong, please correct me if necessary :D

Kerwin Hui

Posted on Monday, 12 April, 2004 - 08:24 pm:

You can choose

δ < ε

if you want (this can always be done), but it doesn't matter. What continuity means is that providing you are sufficiently close (this is what the

δ

does) to the point in question, the values of the function are guaranteed to be not more than an arbitrary amount (this is the

ε

) away from the ''true'' value.

Kerwin

Julian Pulman

Posted on Monday, 12 April, 2004 - 08:25 pm:

ε

can be greater to or equal, or even less than

δ

The real point of continuity is that a small change in $x$ means a small(ish) change in $f (x)$ .

As an example, consider $f (x) = x$ and choose an $ε > 0$

Suppose $| x - y | < δ$ , then $| f (x) - f (y) | = | x - y | < δ$

So if $δ = ε$ then we have found (for an epsilon) a delta for which a small change in $x$ implies a small change in $f (x)$ .

But, we could have had $δ = ε / 1000$

since $| x - y | < δ = ε / 1000$ implies that $| f (x) - f (y) | = | x - y | < ε / 1000 < ε$

So in this sense, think of $δ$ to be a function of $ε$

Amy Sherfield

Posted on Wednesday, 14 April, 2004 - 12:08 am:

Thanx guys

There is just one last question that is bugging me.

Let f(x)={e^-1/|x| if x =/= 0 and 0 if x=0}

Prove that f(x) is continuous in

.

I have been able to prove similar piecewise questions but I am finding this question difficult. Can anyone help??

AMY

David Loeffler

Posted on Wednesday, 14 April, 2004 - 01:00 pm:

Do you see why you only have to worry about x = 0? Now think about what 1/|x| and e^-1/|x| look like as you get closer to 0. Then it should be clear to you how to make the result rigorous.

David

Amy Sherfield

Posted on Friday, 16 April, 2004 - 03:05 pm:

David, I don't see why we only have to worry about

x = 0

. Surely to show

f (x)

is continuous at

a

(where

a = R

f (a)

must be defined and

f (x)

tends to

f (a)

x

tends to

a

, i.e.

δ > 0

then

| x - a | < δ \Rightarrow | f (x) - f (a) | < ε

.

Can someone help me further please.

David Loeffler

Posted on Saturday, 17 April, 2004 - 08:18 pm:

Yes, but there are some standard theorems that can help you here:
- if f is continuous at x and g is continuous at f(x), g(f(.)) is continuous at x
- if f is continuous at x and f(x) =/= 0, 1/f(x) is cts at x
- the exponential function and the absolute value function are continuous everywhere

Any analysis lecture course or textbook should prove these, or at least leave you in a position where you can prove them.

Can you see how to put these together to get a proof that e^-1/|x| is continuous at any given x =/= 0? This frees you up to worry about the awkward point x = 0.

David

David Loeffler

Posted on Saturday, 17 April, 2004 - 08:19 pm:

Just a clarification: by g(f(.)) I mean the function taking x to g(f(x)); this is quite a common notation, and I didn't want to use x to mean two different things in one sentence.

Amy Sherfield

Posted on Sunday, 18 April, 2004 - 04:51 pm:

From your post David,
As we know that e^x and |x| are continuous everywhere, then,
|x| is continuous at a =/= 0 so 1/|x| is continuous at a with |x| =/= 0.
Now can we say that if f=1/|x| continuous at a and e^x is continuous at f(a) then g(f(x))=e^1/|x| is continuous at a.

Is this correct as a proof of e^1/|x| continuous at a =/= 0?

Now what do we need to do at the point x=0?

Amy