Epsilon delta limits.

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  • #26
CompuChip
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Also try going through some of the solutions to problems in the link you gave.
 
  • #27
Borek
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Yes dividing by 0 is all you need to do to make a function undefined at a point. Removeable singularities are good examples because they illustrate the point without being overly pathological.

Your "obvious" reasoning is wrong, [tex]f(x) \neq x + 1[/tex] in order to make your simplification you implicitly assumed that [tex]x\neq 1[/tex], and that's the entire point isn't it! Of course you lose the subtlety if you do bad math! lol

Good point, my bad. Nice trick - take any function, multiply it by (x-1)/(x-1) to get function that is otherwise identical (or am I loosing other fine details now?), but undefined at x=1.
 
  • #28
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If I'm understanding this defintion right, it implies continuity (or defines it...) I'mprobably overthinking it if I'm making conclusions like that. I'll set it asides and try to figure it out another time.

Anyway, many thanks for the attempts at explaining.
 
  • #30
CompuChip
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If I'm understanding this defintion right, it implies continuity (or defines it...) I'mprobably overthinking it if I'm making conclusions like that. I'll set it asides and try to figure it out another time.

Anyway, many thanks for the attempts at explaining.

Continuity of a function f at a just means that the limit
[tex]\lim_{x \to a} f(x) = L[/tex]
exists as in the definition of limit, but moreover that f is actually defined at a and that [itex]L = f(a)[/itex].

So in summary: f is continuous at a means that [tex]\lim_{x \to a} f(x) = f(a)[/tex]. f is continuous means that it is continuous at a for all a in the domain.

But for non-continuous functions, the limit may still exist although the function value is different or not defined. For example,
[tex]f(x) = \begin{cases} x^2 & \text{ if } x \neq 1 \\ \sqrt{2}\pi & \text{ if } x = 1 \end{cases}[/tex]
is not a continuous function, nevertheless the limit as x goes to 1 does exist (and is equal to 1, which is not f(1)).
 
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  • #31
Trambolin, for the first limit you wrote, you set the limit as x approaches infinity. The epsilon-delta proofs for these are a little different than the standard.

For the second and third limits, you must first define a point a that x approaches.

In general, here's how you should approach it:

1. First, find a function f(x) and a point a, which is the point that you want x to approach.
2. Evaluate the limit - your result is L.
3. Set up the inequalities |f(x)-L|<[tex]\epsilon[/tex] and 0<|x-a|<[tex]\delta[/tex].
4. Manipulate both inequalities so that you have the same terms inside the absolute value signs in both inequalities.
5. Set [tex]\delta[/tex]=g([tex]\epsilon[/tex]), and from this, you can find a delta given any epsilon.
 
  • #32
Here's one easy example: Prove that the limit of 3x - 5 as x[tex]\rightarrow[/tex]2 is 1.

Scratch work:
1. Our f(x) = 3x - 5, and a = 2.
2. It is easy to see, either by the problem statement or by evaluating the limit, that L is indeed one.
3. |f(x) - L|< ε if 0<|x - a|< δ, or |(3x-5) -1|< ε if |x - 2|< δ
4. 3|x - 2|< ε if |x - 2|< δ, which is equivalent to |x - 2|< ε/3 if |x - 2|< δ
5. Since the same terms are now inside both absolute value signs, we can see that we must set δ = ε/3. In other words, for whatever ε we pick, δ will be one-third of that number.

Formal proof:
Let δ = ε/3.
|x - 2|< δ = ε/3 if 3|x - 2|< ε, or |3x - 6|< ε. Thus, |(3x - 5) - 1|< ε.
We can see that 0<|x - 2|< δ implies |(3x - 5) - 1|< ε, or equivalently, 0<|x - a|< δ implies |f(x) - L|< ε. QED.

Trambolin, if you were to give me any positive value of ε, we could now find a δ that would satisfy the proof by using δ = ε/3. Also, as you can see, the proof is basically just those 5 steps done backwards.
 
  • #33
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Well actually, if you read carefully, it wasn't me who asked the question... I was trying to tickle the motivation channels of the original poster, and the limits (if exist) are obvious from the simple functions that I specifically choose, but anyway thanks for the clarification.
 
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  • #34
CompuChip
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Thanks for the contribution adartsesirhc. But I hope you did notice that this thread is like, one month old, and post you are referring to is on page 1 of (currently) 3.
 
  • #35
I'm sorry. I only meant to help. The second post was only to clear up any misconception as to the usage of the epsilon-delta proofs.
 
  • #36
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delta epsilon explanation

I can't get my head around the epsilon-delta definition of a limit. Unfortunately I don't have a teacher to ask (I'm teaching this to myself as a self interest) so this forum is my last resort -- google hasn't been kind to me.

From what I've seen, I don't really understand how the definition means much of anything (visual examples included.) All it seems to say is "there is an unspecified value which is greater than the difference between a function and a given value L, where that difference is greater than zero" The problem is, I don't see any need for L or f(x) to be anywhere near one another.

There are three things I would like to say. First, your concern over your lack of understanding of the delta epsilon definition of a limit shows that you have significant mathematical talent.

Second, here is the way I like to phrase the delta-epsilon definition:

DEFINITION: "A function f:R->R aproaches a limit L as x goes to c if and only if
For all epsilon > 0, there exists a delta > 0 such that |x-c| < delta implies that |f(x)-L| < epsilon."

Note that I didn't say anything about "an unspecified value of epsilon." I said, "for ALL positive values of epsilon.

It is an implication of the definition that it is only small values of epsilon that really count. This is because if you can find a "delta," call it delta1, for some small value of epsilon, call it epsilon1, then that value of delta will work for all epsilon greater than epsilon1.

In other words, if you are trying to apply the definition to prove that some number L is a limit as x -> c of some function f(x) and you suceed in finding a delta (delta1) that works for some value of epsilon (epsilon1), then that value of delta (delta1) will work for all epsilon2 > epsilon1. Hence you are finished with values of epsilon greater than epsilon1 and all you have to worry about now is the values of epsilon less than epsilon1.

One could state this as a Theorem:

Suppose that there exists epsilon1, delta1 > zero such that |f(x)-L| < epsilon1 for all x such that |x-c| < delta1, THEN for all epsilon2 > epsilon1 it is true that |f(x)-L| < epsilon2.

Proof: |f(x)-L| < epsilon1 < epsilon2. QED.

The third thing I would like to say is that for the three months of summer vacation before I went to college 45 years ago, I spent a lot of time trying to understand the delta-epsilon definition of a limit and of a derivative. I didn't feel I understood it after those three months. But, a couple of weeks into my freshman calculus class, I felt I really understood it. I don't remember much else about what happened back then, but, I do remember that the Professor invited me to switch from the regular calculus class to the honors calculus class after the first month of class. I thanked him and said that I was learning too much in the regular class to switch, but, could I take both? He said "of course." The point is that the manyt many hours over many many months of thinking about that definition served me well not only as an undergraduate but also as a graduate student. All phases of analysis (indlucing PDE, functional analysis, and C* algebras) were vastly much easier for me because I learned that definition. Same thing with mathematical logic and general topology.

Algebra and algebraic topology and algebraic geometry was difficult for me, though, and still is. Not to mention number theory. I still don't have my belt all the way around Chern classes.

Deacon John
 
  • #37
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Deacon John, can you tell me please using your definition of limit what IS the following limit.
1) lim 2x+1 as x-------> 2 where f:N------>R where N is the natural Nos and f(x)=2x+1
Does the following function has any limits within its domain and if yes how many???
f={(1/n,1+2^-n):nεΝ}
 
  • #38
HallsofIvy
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What is your motivation in asking these questions? They seem fairly simple to me and Deacon John's definition of limit gives the same thing as the "usual" limit.

For f(x)= 2x+ 1, as x goes to 2, f:N->R, the limit, by Deacon John's definition, or any definition I am familiar with is 5: Since x is an integer, the only way we can have |x- 2|< [itex]\delta[/itex] for small delta is to have x= 2 so the limit is just f(2)= 5.

For the second question, again since to be "close" to an integer, n must be that integer, every point in the range of the function is a limit point and, since n can be arbitrarily large, (0, 1) is also a limit point. That function has a countably infinite number of limit points.

The only difficulty I can see with Deacon John's definition of limit is that it does not include 0< |x-c|< delta rather than just |x-c|< delta. (And I always forget that myself!)

To point that out, you might ask "what is the limit of the function 'f(x)= x if x is NOT 0, f(0)= 1' as x goes to 0? (f:R->R)"

Using the strict wording of Deacon John's definition, that has NO limit since for x arbitrarily close to 0, f(x) takes on values arbitrarily close to 0 AND 1. Excluding |x-c|= 0, that is excluding x= 0 in this case, f(x) takes values arbitrarily close to 0 only and the limit is 0. (That difference is crucially important to the definition of the derivative- where we are always taking the limit of a fraction which is undefined at the limit value.)
 
  • #39
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Deacon John, can you tell me please using your definition of limit what IS the following limit.
1) lim 2x+1 as x-------> 2 where f:N------>R where N is the natural Nos and f(x)=2x+1
Does the following function has any limits within its domain and if yes how many???
f={(1/n,1+2^-n):nεΝ}

Lavranos,

The definition that I gave was only for functions that map the real numbers into the real numbers.

The definition that I gave does not apply when the domain of the function is the natural numbers (N).

When the domain is the natural numbers, there are a whole host of definitions that are possible.

[For the advanced student: To see this, pick any point in the Stone Chech compactification of N (call the point omega), pick any point of N, for example the number "3," and use your favoite method to identify "3" with "omega." If this is not immediately clear, you might not be an advanced student, but, don't be discouraged, everybody was a beginner when they started out. I pretty sure that different points give different topologies. For example, I'm pretty sure that the S.C.c. contains a point in the closure of the even numbers but not in the closure of the odd numbers.]

Probably the most natural definition of a limit for your kind of functions is given by the discrete topology, where every subset of natual numbers is both closed and open. To apply this requires a more sophisticated definition of a limit than I am willing to explain, but the result is that every function is continuous and every value in the domain (N) and every value in the image (i.e., the set f(N)) of every function (defined on N, that is) is a limit.

The answer to your first question (assuming the discrete topology on N) is "5" because f(2) = 5. And, with the same assumption, every natural number is a "limit" in your sense of the word, i.e., a point in the domain of the function, where the function approaches a limit.

Howeever, your second example is more likely to be associated with questions about the limit points of the image of the second function.

You did not specify the range for your second function, but the most natural choice would be the Cartesian product of the real numbers with themsleves, namely, RxR. I will assume that this is what you meant.

Your second function has the point (0,1) as a limit point of it's image, but this point is not in the image of the second function.

In fact, the limit as x -> infinity of your second function is (0,1), and "infinity" is not in the domain of your second function.

I did not give a "delta - epsilon" definition for this kind of limit, but, there is one. I suggest you look it up. It should be in the same book where you find the one that I did give.

Or better, use your intuition to write down what it should look like, then look it up.

Hint: The "delta-epsilon" definition for the kind of of limit when n --> infinity does not have any "delta" in it, but, it does have an "epsilon."

[For the advanced student: the definition of "limit" that I am trying to motivate Lavranos to "discover" is equivalent to the definition when the topology of the "one point compactification" is put on the set N+ = N union {infinity}. I.e., the closed subsets of N+ are precisely the finite subsets and N+ itslef. As far as I know, "N+" is not a standard notation.]

Cheers,

DJ
 
  • #40
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THE FUNCTION f(x)=2x+1 where f:N------->R HAS NO LIMIT AT X=2 BECAUSE X IS NOT A POINT OF ACCUMULATION ,BUT IT IS CONTINUOUS AT X=2.
THE SECOND FUNCTION IS NOT MINE IT IS A PROFESSOR"S FUNCTION AND HE SAYS THAT THE FUNCTION CAN HAVE ALIMIT AT X=0 which is 1,and he proves that.
I SUGGEST you write my post on apiece of paper to make sure you don't make any mistakes and go and check it with couple of books.
But then again the question comes up :DO YOU KNOW HOW TO READ BOOKS
PROPERLY????
Deacon John the above is not for you ,please go and see my posts in the thread <Discussion of valid method of proof> under General maths and i will ask a few more questions.
 

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