# Epsilon delta limits.

kts123
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.

For example: f(5)==>25 ;

limx=>6 f(x) = 20

-1000 < |f(5) - 20| < 1000

And

0 < |5-6| < 2

That is, I see no specific reason to even bother putting in the correct values. How does the definition define anything if we require prior knowledge of what our limits and x values are? Furthermore, how are epsilon and delta even related? The visual explanations make even less sense -- why can't we choose two given values of f(x) and L to which the difference is large, yet select a large enough epsilon so that the centre of f(x) and L lies between the bands of epsilon.

I'm terribly confused as everyone can probably tell... sorry.

csprof2000
All it is meant to do is provide a rigorous justification for doing limits. I mean, it's on the one hand just some silly formalism, and on the other hand it does make rigorous proofs more straightforward and feasible.

The idea is that, if a limit exists, there is an interval (say, of width delta) around the point in which the function is close (say, within a small difference epsilon) to some value, lim f(x). Doing delta-epsilon proofs is just a rigorous way of showing that this interval exists, and thus the limit exists.

It's a bit awkward, but go ahead and give it a shot. Don't beat yourself up over it.

Homework Helper
imagine a function which is monotonically decreasing, and takes on all positive values less than 1, as x approaches 4.

can you see why, even prove , its limit is zero as x goes to 4?

kts123
I can't help but get a tingly feeling on my tongue that there's something very profound behind this definition. Unfortunately I'll have to beat myself up over it for a few months before I give up, that's how it always is. =(

kts123
ALL positive values of less than 1 as x goes to 4? I suppose the smallest positive value less than one is greater than zero by the smallest possible amount, leaving the only value "less than" to be zero itself (that is to say, since "A" is the smallest possible value greater than zero, and since f(x) > f(x+B) (where B is any positive number, since of course the function is always decreasing.) Thus when f(x)==>A, then f(x+B)<f(x), and since the only value less than A is zero itself...

Errr... I think that makes sense.

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Homework Helper
Actually, limits are a very intuitive concept. I think you (and any starting mathematician, even) would agree with me that
by a statement like $$\lim_{x \to a} f(x) = L$$ we mean that the closer x gets to a, the closer the function value f(x) gets to L.​
In other words: if we look at the function value in a point close enough to a, we are certain to find something which is close enough to L. The epsilon-delta stuff is just a way to rigorously define what we mean by "close enough".

It works as follows:
• Let $$\epsilon > 0$$.
• Then there exists a $$\delta > 0$$ such that: if $$|x - a| < \delta$$, then $$|f(x) - L| < \epsilon$$.
The first point says, you can choose epsilon. This epsilon will be the maximum allowed difference between the function value and the limit value. You can make it as small or as large as you want. If you keep making it smaller and smaller, the function value will be closer and closer to the limit. Note that it does explicitly say that epsilon is positive: the statement does not say that there will be a point where epsilon is actually zero (that is, the function actually reaches the limit).
Then the second point asserts the existence of a neighborhood of size delta around a for which you will find the function value within epsilon of L. That is, whatever value of x in the interval $(a - \delta, a + \delta)$ I plug into f, I am sure that the function value will be in the interval $(L - \epsilon, L + \epsilon)$. Of course, the smaller the epsilon you give me (the more stringent your definition of what you call "close enough") the smaller my delta will have to be (the closer you will have to be to a to actually come "close enough").

Mentor
How does the definition define anything if we require prior knowledge of what our limits and x values are?

Definition is not a recipe for limit calculation. However, it tells you how to check if the number you suspect to be the limit is the limit - or not.

kts123
Ohhhh, thanks CompuChip, I was expecting there to be something that implied "$$\lim_{\epsilon \to 0}$$" (that is, something which forced epsilon to squash f(x) and L together.) I have to wonder, though, isn't that definition somewhat useless for a function which oscillates extremely rapidly near the limit; for example, sin(1/$$x^2$$), where x is very small, say 2E(-10000)?

Definition is not a recipe for limit calculation. However, it tells you how to check if the number you suspect to be the limit is the limit - or not.

Amazing that no sites wanted to mention these things. I suppose this is a testimate to textbooks being much more reliable than internet sources. =P

Homework Helper
I have to wonder, though, isn't that definition somewhat useless for a function which oscillates extremely rapidly near the limit; for example, sin(1/$$x^2$$), where x is very small, say 2E(-10000)?
No, in fact it is extremely useful because it allows you to prove that
$$\lim_{x \to 0} \sin(1 / x^2)$$
does not exist (which you of course already suspected by a reasoning like: "if x goes to zero, 1/x^2 goes to infinity, and the sine function keeps oscillating there, so probably there is no limit"). For suppose you claim that the limit is $-1 \le L \le 1$, then I can show that there is an epsilon such that for all delta, there is an x closer than delta to zero for which sin(1/x^2) will be more than epsilon from L. Which by the very definition says: "the limit is not L".

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Mentor
Amazing that no sites wanted to mention these things. I suppose this is a testimate to textbooks being much more reliable than internet sources.

I am afraid textbooks will be no better here. Many math books leave lots of things untold - as these are obvious. Fact that these things are obvious only for mathematicians, doesn't matter - at least to them

Homework Helper

Ohhhh, thanks CompuChip, I was expecting there to be something that implied "$$\lim_{\epsilon \to 0}$$" (that is, something which forced epsilon to squash f(x) and L together.)

Note that there is no such thing explicitly. But of course, they idea is that you will make epsilon smaller and smaller. You are allowed to choose $\epsilon = 10^{10^{100}}$, and the statement should still be true (and in fact, chances are it is, unless you have chosen the suspected limit really badly or you have a strange function). The point is, however, that it holds for all $\epsilon > 0$, no matter how small. So if there is a number for which it is not true, whether it be very large or very small, then the definition is not satisfied. In fact, as I already stated implicitly in my last post, the opposite of
$$\lim_{x \to a} f(x) = L$$, meaning: for each $\epsilon > 0$ there is a $\delta = \delta(\epsilon) > 0$ such that for all x satisfying $|x - a| < \delta$, $|f(x) - L| < \epsilon$​
is
$$\lim_{x \to a} f(x) \neq L$$, meaning: there is an $\epsilon > 0$ (finding just one suffices) such that for all $\delta = \delta(\epsilon) > 0$ there is an x satisfying $|x - a| < \delta$ for which $|f(x) - L| \ge \epsilon$​
Which is: there is a distance $\epsilon$ for which I cannot get f(x) closer than $\epsilon$ to the supposed limit L, whence it cannot be the limit.

kts123
No, in fact it is extremely useful because it allows you to prove that
$$\lim_{x \to 0} \sin(1 / x^2)$$
does not exist (which you of course already suspected by a reasoning like: "if x goes to zero, 1/x^2 goes to infinity, and the sine function keeps oscillating there, so probably there is no limit"). For suppose you claim that the limit is $-1 \le L \le 1$, then I can show that there is an epsilon such that for all delta, there is an x closer than delta to zero for which sin(1/x^2) will be more than epsilon from L. Which by the very definition says: "the limit is not L".

I meant when x is very small, but not EXACTLY zero. For examle, $$\lim_{x \to 10^{-10000}} \sin(1/x^2)$$ The required value for epsilon and delta is so incredibly small that it seems rather difficult to "pick a value that 'seems close enough," it would be very easy to accidently pick a value too large or too small. I suppose I was hoping the formal definition would introduce something less... ambiguous (though I suppose it needs to be based on the nature of a limit.) It is, as you say, very useful, just not the solution to all your limit needs. Anyway, many thanks for the advice, everyone.

Homework Helper
I don't really see what your problem with that definition is. There is no "required value" of epsilon, you can choose it, as I tried to explain. For any given epsilon, finding the right value of delta can indeed be a tough job, in fact, that's the hardest part about epsilon-delta proofs.

Also, the "problem" with the function you introduced only occurs at x = 0, where the function indeed has no limit (as you can actually prove from the definition). You can also prove from the definition that $$\lim_{x \to 10^{-10000}} \sin(1/x^2) = \left. \sin(1 / x^2) \right|_{x = 10^{-10000}}$$
(or you can prove first that if f is continuous in a, then $\lim_{x \to a} f(x) = f(a)$). It is not any easier or more difficult than proving that $$\lim_{x \to 1} \sin(1 / x^2) = \sin(1)$$. In general in such proofs, you start by writing down "Let $\epsilon > 0$" and then proceed to define delta in terms of epsilon, e.g. "Choose $\delta = \sqrt{\epsilon} / 4$ and let x be such that $|x - a| < \delta$." Then you go and estimate the difference f(x) - L in a series of inequalities to show that it is smaller than epsilon. Whether a is 0 or infinity, or something in between, is irrelevant for that structure of the proof and does not increase or decrease the level of difficulty.

Can you explain what it is that you find ambiguous about the definition?

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trambolin
The simpler, the better...

Find the limit of f(x) = x when x goes to infinity, I choose you epsilon = 17 and 3506, can you give me a delta such that the conditions satisfied?

What about f(x) = 1/x? I give you epsilon = 1849 and 0.02? Any deltas?

And lastly, f(x) = cos(x), I choose epsilon = 0.5...

kts123
I don't really see what your problem with that definition is. There is no "required value" of epsilon, you can choose it, as I tried to explain. For any given epsilon, finding the right value of delta can indeed be a tough job, in fact, that's the hardest part about epsilon-delta proofs.

Also, the "problem" with the function you introduced only occurs at x = 0, where the function indeed has no limit (as you can actually prove from the definition). You can also prove from the definition that $$\lim_{x \to 10^{-10000}} \sin(1/x^2) = \left. \sin(1 / x^2) \right|_{x = 10^{-10000}}$$
(or you can prove first that if f is continuous in a, then $\lim_{x \to a} f(x) = f(a)$). It is not any easier or more difficult than proving that $$\lim_{x \to 1} \sin(1 / x^2) = \sin(1)$$. In general in such proofs, you start by writing down "Let $\epsilon > 0$" and then proceed to define delta in terms of epsilon, e.g. "Choose $\delta = \sqrt{\epsilon} / 4$ and let x be such that $|x - a| < \delta$." Then you go and estimate the difference f(x) - L in a series of inequalities to show that it is smaller than epsilon. Whether a is 0 or infinity, or something in between, is irrelevant for that structure of the proof and does not increase or decrease the level of difficulty.

Can you explain what it is that you find ambiguous about the definition?

That the definition doesn't say what value epsilon or delta should be. By the second part we say $0 < |x - a| < \delta$. That's saying that no matter how small we make delta, the difference of x and a will always be smaller -- but not fully zero? It's SMALLER than ANY number you could EVER choose, but NOT zero? What the heck is the reciprocal of such a small number? BIGGER than any number you could choose? I understand what the definition is saying, it just seems weird.

Mentor
That the definition doesn't say what value epsilon or delta should be.

Because it doesn't matter what values they have, as long as they fulfill the condition.

Homework Helper
That the definition doesn't say what value epsilon or delta should be.
That's because epsilon can be anything (as I explained), and delta depends on epsilon and the function itself. The definition works for any function, exactly because it doesn't prescribe delta.

By the second part we say $0 < |x - a| < \delta$. That's saying that no matter how small we make delta, the difference of x and a will always be smaller -- but not fully zero? It's SMALLER than ANY number you could EVER choose, but NOT zero?
Sorry, I missed the part where it says that |x - a| cannot be zero. The definition says: if |x - a| < delta, then |f(x) - L| < epsilon. In particular, you can do it for x = a, if the function is defined there (i.e. for a continuous function). But for functions like 1/x you cannot make |x - 0| equal to 0 because that would force you to choose x = 0 but the function is not defined there.

What the heck is the reciprocal of such a small number? BIGGER than any number you could choose?
In the phrase "such a small number", which number are you talking about? Epsilon? Delta?
And the reciprocal of a small number is indeed a large number, but what does that have to do with it?

I understand what the definition is saying, it just seems weird.
Actually, once you understand what the definition is trying to express, I think it's a completely logical way to do it, and one of the few ways that expresses the idea of a limit without making use of that concept. So if you don't mind me being very blunt, I don't think you fully understand it yet

kts123
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DavidWhitbeck
Yeah that's right you don't consider x = a in the definition. The limit can exist even if the function is not defined at that point (which would not be true if you needed to include x=a in the definition). For example $$f(x) = \frac{x^2-1}{x - 1}$$ is not defined at x=1 but $$\lim_{x\rightarrow 1} f(x) = 2$$.

Homework Helper
Yeah that's right you don't consider x = a in the definition. The limit can exist even if the function is not defined at that point (which would not be true if you needed to include x=a in the definition).

Infact the values of f only need be considered when 0<|x-delta|<epsilon for any epsilon
no only can the function be undifined elsewhere it may be differntly defined the limit checks if the function is almost constant in any small neighborhood of the value

Mentor
For example $$f(x) = \frac{x^2-1}{x - 1}$$ is not defined at x=1

I understand what you are aiming at, I just wonder if it is a good example, as obviously

$$f(x) = \frac{x^2-1}{x - 1} = \frac{(x+1)(x-1)}{x - 1} = x + 1$$

So the question is, is multiplying by 0/0 enough to make function undefined in a point?

kts123
Yeah that's right you don't consider x = a in the definition. The limit can exist even if the function is not defined at that point (which would not be true if you needed to include x=a in the definition). For example $$f(x) = \frac{x^2-1}{x - 1}$$ is not defined at x=1 but $$\lim_{x\rightarrow 1} f(x) = 2$$.

I know what you're getting at, I've delt with a lot of limits despite not knowing the formal definition. What I don't understand is that it says, in otherwords, $$0<|x-a|<\delta$$, where delta is ANY real number greater than zero. No matter how small we make delta, |x-a| is still smaller -- but not zero. I don't see anywhere that says "the definition only counts if you actually choose delta," by all rights, delta is always defined for ANY value greater than zero, even if we don't "select" delta. Essentially, it's sayings |x-a| is positive, yet smaller than any positive number that exists.

kts123
Yeah that's right you don't consider x = a in the definition. The limit can exist even if the function is not defined at that point (which would not be true if you needed to include x=a in the definition). For example $$f(x) = \frac{x^2-1}{x - 1}$$ is not defined at x=1 but $$\lim_{x\rightarrow 1} f(x) = 2$$.

I know what you're getting at, I've delt with a lot of limits despite not knowing the formal definition. What I don't understand is that it says, in otherwords, $$0<|x-a|<\delta$$, where delta is ANY real number greater than zero. No matter how small we make delta, |x-a| is still smaller -- but not zero. I don't see anywhere that says "the definition only counts if you actually choose delta," by all rights, delta is always defined for ANY value greater than zero, even if we don't "select" delta. Essentially, it's sayings |x-a| is greater than zero, yet smaller than any other number greater than zero.

DavidWhitbeck
I understand what you are aiming at, I just wonder if it is a good example, as obviously

$$f(x) = \frac{x^2-1}{x - 1} = \frac{(x+1)(x-1)}{x - 1} = x + 1$$

So the question is, is multiplying by 0/0 enough to make function undefined in a point?

Borek

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, $$f(x) \neq x + 1$$ in order to make your simplification you implicitly assumed that $$x\neq 1$$, and that's the entire point isn't it! Of course you lose the subtlety if you do bad math! lol

DavidWhitbeck
I know what you're getting at, I've delt with a lot of limits despite not knowing the formal definition. What I don't understand is that it says, in otherwords, $$0<|x-a|<\delta$$, where delta is ANY real number greater than zero. No matter how small we make delta, |x-a| is still smaller -- but not zero. I don't see anywhere that says "the definition only counts if you actually choose delta," by all rights, delta is always defined for ANY value greater than zero, even if we don't "select" delta. Essentially, it's sayings |x-a| is positive, yet smaller than any positive number that exists.

You don't actually have to choose delta, but you must show that it exists. Let's go over this as a recipe. Let's suppose you think the limit exists and equals the real number L.

(1) Choose $$\epsilon > 0$$
(2) We need to find a real number $$\delta$$ such that whenever $$|x - a| < \delta$$, the following inequality is also satisfied $$|f(x) - L| < \epsilon$$
(3) If we can repeat (1)-(2) successfully for all positive values of $$\epsilon$$ then we can conclude that the limit exists.

Now look at that procedure. It is very easy to do for all those cases at once if you find delta as a function of epsilon, $$\delta = g(\epsilon)$$. You should think of it that way-- from the epsilon you find delta, which in turn gives you the inequality for delta that must imply the inequality for epsilon. It's like a big circle.

By saying that you think delta is always defined, you clearly don't understand the definition yet. You, yourself must find delta to show that the limit exists. It's not a priori defined.

Homework Helper
Mentor
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, $$f(x) \neq x + 1$$ in order to make your simplification you implicitly assumed that $$x\neq 1$$, 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.

kts123
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.

Edgardo
Homework Helper
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
$$\lim_{x \to a} f(x) = L$$
exists as in the definition of limit, but moreover that f is actually defined at a and that $L = f(a)$.

So in summary: f is continuous at a means that $$\lim_{x \to a} f(x) = f(a)$$. 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,
$$f(x) = \begin{cases} x^2 & \text{ if } x \neq 1 \\ \sqrt{2}\pi & \text{ if } x = 1 \end{cases}$$
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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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|<$$\epsilon$$ and 0<|x-a|<$$\delta$$.
4. Manipulate both inequalities so that you have the same terms inside the absolute value signs in both inequalities.
5. Set $$\delta$$=g($$\epsilon$$), and from this, you can find a delta given any epsilon.

Here's one easy example: Prove that the limit of 3x - 5 as x$$\rightarrow$$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.

trambolin
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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