| 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 e > 0, there exists a d > 0 such that |x-a| < d Þ |f(x)-f(a)| < e.'' If I had to prove that a function is continuous at a particular point, e.g. x2 at 0, for example, I know that d, e > 0 but can e > d or e = d or e < d. Which of these possibilities can occur for e and d from the definition I have written in quotes?
Thanx
AMY
| 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)=x2 at x=0, choose
| d = | Ö |
e |
. Then if |0-a| < d a < d a2 < e |02-a2| < e |f(0)-f(a)| < e
therefore continuous at x=0
-Phil.
pps. dear uni guys: never done that before, probably very wrong, please correct me if necessary :D
| Posted on Monday, 12 April, 2004 - 08:24 pm: |
You can choose d < e 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 d does) to the point in question, the values of the function are guaranteed to be not more than an arbitrary amount (this is the e) away from the ''true'' value. Kerwin
| Posted on Monday, 12 April, 2004 - 08:25 pm: |
e can be greater to or equal, or even less than d. 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 e > 0 Suppose |x-y| < d, then |f(x)-f(y)|=|x-y| < d So if d = e 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 d = e/1000 since |x-y| < d = e/1000 implies that |f(x)-f(y)|=|x-y| < e/1000 < e So in this sense, think of d to be a function of e
| 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
| 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
| 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) as x tends to a, i.e. d > 0 then |x-a| < dÞ |f(x)-f(a)| < e.
Can someone help me further please.
| 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
| 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.
| Posted on Sunday, 18 April, 2004 - 04:51 pm: |
From your post David,
As we know that ex 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 ex is continuous at f(a) then g(f(x))=e1/|x| is continuous at a.
Is this correct as a proof of e1/|x| continuous at a =/= 0?
Now what do we need to do at the point x=0?
Amy