Question

Delta Epsilon Proof Tutorial (Advanced Level)
This tutorial will require you to have a good understanding of algebra. I believe this tutorial will be helpful to tutors more so than students, since many people on OS are unfamiliar or weary of Delta Epsilon.

Answer 1

@hartnn

There are some typos, but I think it should help quite a bit.
Let me know if there are any confusing parts.

Answer 2

I'm going to fix some of the typos now.

Answer 3

\(\Large \text{The Delta Epsilon Definition}\)
The following limit: \[
\lim_{x\to a}f(x)=L
\]Is defined as:
For any \(\epsilon\gt0\), there must be a \(\delta\gt0\) such that:
If \(x\neq a\) and \(a-\delta \lt x \lt a+\delta \) then \(L-\epsilon \lt f(x) \lt L+\epsilon\).
We we often will write this as: \[
\forall \epsilon\gt0\exists\delta\gt0\forall x \quad0\lt|x-a|\lt\delta \implies |f(x)-L|\lt\epsilon
\]Understand that \(\epsilon\) represents how far \(f(x)\) is allowed to stray from \(L\), while \(\delta\) represents how far \(x\) is allowed to stray from \(a\).

Also notice that \(0\lt|x-a|\), which means that \(x\neq a\).
This is actually the main reason we use limits. Limits allow us to operate under the assumption that \(x\neq a\). This means that \(f(a)\) does not need to be defined, it just need to be an indeterminate form.

Answer 4

\(\Large \text{The Delta Epsilon Game}\)
Think of it as a game we are playing against our opponent. Our opponent is just not convinced and will do whatever is possible to find a counter example. It isn't enough to put the opponent in check, we must put them in checkmate to win this game. Like any other proof, there can't be any counter examples or holes.

Notice the order of the qualifiers. \[
\forall \epsilon\exists\delta\forall x \quad0\lt|x-a|\lt\delta \implies |f(x)-L|\lt\epsilon
\]Each qualifier is a step in our game.

1. The opponent chooses \(\epsilon \).
2. We choose \(\delta\).
3. The opponent chooses \(x\).

The order matters because it determines what each player knows.
When the opponent choose \(\epsilon\), he doesn't know anything about \(\delta\) or \(x\).
When we choose \(\delta\), we know \(\epsilon\) but not \(x\).
When the opponent choose \(x\), they know about \(\delta\) and \(\epsilon\).

As a consequence of all this \(\delta\) cannot depend on \(x\) since it is not known what \(x\) is when we are made to chose \(\delta\). When we choose \(\delta\), it must depend entirely on \(\epsilon\). So ultimately \(\delta\) must be a function of \(\epsilon\).

\(\Large \text{The Optimal Strategy}\)
The ultimate strategy is to let \(\delta\) to be as small as possible.
If we are ever given a choice between \(h\) and \(k\), we want to chose \(\delta=\min(h,k)\).
Why is this? The reason is that the smaller we let \(\delta\) be, the less \(x\) values we have, which means that there are less \(f(x)\) values, and less chances of \(|f(x)-L|\ge \epsilon\).
Remember how conditional statements work. \(a\implies b\) is false when \(a\) is true and \(b\) is false. If \(|x-a|<\delta\) and \(|f(x)-L|\ge \epsilon\) for any \(\epsilon\), then we lose the game.

Unfortunately, we are stuck with the restriction \(0\lt\delta\).
So what is the smallest possible \(\delta\) we could choose? Well that is like asking what is the largest number. You can always make a number bigger by adding \(1\), and you can always find a smaller positive number by dividing by \(2\).

In order to win this games, we need to find a function \(\delta=g(\epsilon)\), which will work regardless of the \(\epsilon\) value thrown at us. Fortunately our function can use other functions such as \(\min\).

Answer 5

\(\Large \text{Delta Epsilon and Polynomials and Rational functions}\)
Usually Calculus students will be asked to do delta epsilon proofs for polynomials and rational functions.
Here is an algorithm that will always work for polynomials and rational functions, and may sometimes work for other functions.
Consider the rational function \(f(x)\) with the following limit:\[
\lim_{x\to a}f(x) = L
\]For this limit to exist, it must be the case that \((x-a)\) is a factor of \(f(x)-L\).

Let's look at what our definition will look like:\[
0\lt|x-a|\lt \delta \implies |f(x)-L|\lt\epsilon
\]Since it is rational function, we will be able to factor out \(x-a\) from the numerator.\[
0\lt|x-a|\lt \delta \implies \left|(x-a)\frac{f(x)-L}{x-a}\right|\lt\epsilon\\
0\lt|x-a|\lt \delta \implies |x-a|\left|\frac{f(x)-L}{x-a}\right|\lt\epsilon
\]Next, we will divide both sides of the \(\epsilon\) inequality of whatever remains after having factored out \(x-a\).\[
0\lt |x-a|\lt\delta \implies |x-a|\lt\epsilon\left|\frac{x-a}{f(x)-L}\right|
\]At this point, we would win the game if we could say that: \[
\delta=g(\epsilon,x)=\epsilon\left|\frac{x-a}{f(x)-L}\right|
\]However, we are not allowed to do this because \(\delta\) must be independent of \(x\).
Remember that we are only given an \(\epsilon\) value but not an \(x\) value.
The good news is that we control what \(x\) can be through \(\delta\).

\(\Large \text{Fail-safe: Delta Subscript Zero}\)
In this case, what we generally do is pick some \(\delta_0\) and let \(\delta =\min(\delta_0,g(\epsilon) )\).
This means \(\delta\leq \delta_0\). Which means: \[
|x-a|\lt\delta\implies |x-a|\lt\delta_0\implies a-\delta_0 \lt x\lt a+\delta_0
\]Next we let: \[
g(\epsilon) = \min_{a-\delta_0\lt x\lt a+\delta_0}\left(\epsilon\left|\frac{x-a}{f(x)-L}\right|\right) = \epsilon \left(\min_{a-\delta_0\lt x\lt a+\delta_0}\left(\left|\frac{x-a}{f(x)-L}\right|\right)\right)
\]All we need to do now is to prove that the minimum does not equal \(0\).

To do that, let's talk a bit about: \[
h(x)=\left|\frac{x-a}{f(x)-L}\right|
\]First of all \(h(x)\geq 0\), so at least we know it won't be negative.
Second of all \(h(x)\) is a rational function. This means that it is going to have a finite number of roots.
Therefore, so long as we pick a \(\delta_0\) such that \(h(x)\neq 0\) for all \(x\in[a-\delta_0,a+\delta_0]\) then we can be certain that: \[
g(\epsilon) = \epsilon \min(h(x))\gt 0
\]
Please note that these proofs will expect you to explicitly pick a \(\delta_0\), and then you will have to actually find \(\min(h(x))\), to get rid of any \(x\) terms.

Answer 6

\(\Large\text{Delta Epsilon Example}\)
I've been very abstract, but it's time for an actual example to help make sense of it all.
Let's prove that: \[
\lim_{x\to 2}\frac{x^2-x-2}{x-2} = 3
\]First let's take note that: \(x^2-x-2 = (x-2)(x+1)\).
We knew we had to have an indeterminate form, so we know that \(x-2\) has to be a factor of the numerator.
Now on to our definition: \[
0\lt|x-2|\lt \delta \implies |f(x)-3|\lt \epsilon\\
0\lt|x-2|\lt \delta \implies \left|\frac{x^2-x-2}{x-2}-3\right|\lt \epsilon \\
0\lt|x-2|\lt \delta \implies \left|\frac{(x-2)(x+1)}{x-2}-3\right|\lt \epsilon \\
0\lt|x-2|\lt \delta \implies |(x+1)-3|\lt \epsilon \\
0\lt|x-2|\lt \delta \implies |x-2|\lt \epsilon
\]Well, this one turned out nice and easy.
All we have to do is let \(\delta = g(\epsilon) =\epsilon\).

Let's to a slightly harder one:\[
\lim_{x\to 2} x^2-x-2= 0
\]
Now on to our definition: \[
0\lt|x-2|\lt \delta \implies |f(x)-0|\lt \epsilon\\
0\lt|x-2|\lt \delta \implies \left|(x^2-x-2)\right|\lt \epsilon \\
0\lt|x-2|\lt \delta \implies |(x-2)(x+1)|\lt \epsilon \\
0\lt|x-2|\lt \delta \implies |x-2||x+1|\lt \epsilon\\
0\lt|x-2|\lt \delta \implies |x-2|\lt \frac{\epsilon}{|x+1|}
\]We want to say: \[
\delta= \frac{\epsilon}{|x+1|}
\]But we can't, so we need to pick a \(\delta_0\).
Fortunately, \(1/|x+1|\) doesn't have any roots to worry about, so we can pick any \(\delta_0\) we want. In these cases, I like to pick \(\delta_0=1\) because it is an easy number to work with. \[
|x-2|\lt\delta\implies |x-2|\lt1\implies -2-1\lt x\lt -2+1 \implies -3\lt x\lt -1
\]Now we minimize \(1/|x+1|\) on the interval \([-3,-1]\): \[
\min\left(\frac{1}{|x+1|}\right) = \frac{1}{\max(|x+1|)} = \frac 12
\]Finally, to finish the proof: \[
\delta = g(\epsilon) = \min\left(\delta_0,\frac{\epsilon}{|x+1|}\right) = \min\left(1,\frac{\epsilon}2\right)
\]

Answer 7

maybe append this with sequences?

Answer 8

It is much nicer, imo, to use the sequential definition of a limit of a function.

Answer 9

I just realized this;)

Answer 10

What do you mean the sequential definition?

Answer 11

I'm using a definition used by text books that actually expect you to do these types of proofs, rather than using limit rules.

Answer 12

not limit rules

Answer 13

there are many definitions that are equivalent, its just easier to work with the sequential definition.

Answer 14

I got a little confused towards the end of your second example

Answer 15

Which part was confusing?

Answer 16

\(\color{blue}{\text{Originally Posted by}}\) @wio

We want to say: \[
\delta= \frac{\epsilon}{|x+1|}
\]But we can't, so we need to pick a \(\delta_0\).
Fortunately, \(1/|x+1|\) doesn't have any roots to worry about, so we can pick any \(\delta_0\) we want. In these cases, I like to pick \(\delta_0=1\) because it is an easy number to work with. \[
|x-2|\lt\delta\implies |x-2|\lt1\implies -2-1\lt x\lt -2+1 \implies -3\lt x\lt -1
\]Now we minimize \(1/|x+1|\) on the interval \([-3,-1]\): \[
\min\left(\frac{1}{|x+1|}\right) = \frac{1}{\max(|x+1|)} = \frac 12
\]Finally, to finish the proof: \[
\delta = g(\epsilon) = \min\left(\delta_0,\frac{\epsilon}{|x+1|}\right) = \min\left(1,\frac{\epsilon}2\right)
\]
\(\color{blue}{\text{End of Quote}}\)
That portion.

Answer 17

lol

Answer 18

I only laugh because this is the most failed class at my school. and you are ttrying to figure it out in 5 mins

Answer 19

I T.A. for this class and 35% passed

Answer 20

I get most of it,I'm not understanding the notation he used to minimize the functino because I learned it a little differently.

Answer 21

Okay, do you understand that we have to eliminate the \(x\)?

Answer 22

Oh, you don't get how I minimized the function?

Answer 23

How you minimized it.

Answer 24

For \(f:D\rightarrow \mathbb{R}\) we have \(\lim_{x\rightarrow x_0}f(x)=L\) provided that if \(x_n\) is a sequence that goes to \(x_0\) on \(D-\{x_0\}\), then \(f(x_n)\rightarrow f(x_0)\)

This is another useful definition, and its exactly the same.

Answer 25

Well, do you understand how \(\max(|x+1|)\) would minimize the inverse?
And you understand that since \(|x+1|\) is just a V shaped line? So it means that it must be maximized at the endpoint of the interval. The endpoints were \(-3\) and \(-1\). Plugging them in get \(2\) and \(0\) respectively. So \(\max(|x+1|)=2\).

Answer 26

That makes more sense.

Answer 27

@zzr0ck3r It's subjective as to what definition is easier. You can't say that your definition is always easier. What really matters here is what was introduced to the student. In many cases it will be Delta Epsilon. You can make your own sequence tutorial if you want. I don't see why I have to throw away Delta Epsilon just because you don't like it.

Answer 28

@Jhannybean I sort of hand waved the \(\min(\ldots)\) part because finding minimums is usually easier than actually doing these proofs.

Answer 29

I see.

Answer 30

But it is also why I say "I believe this tutorial will be helpful to tutors more so than students"

Answer 31

:) Understandable.

Answer 32

We should do a series on these starting at sequences, then continuity of functions, then limits, then differentiation (using the definitions)

I think its a huge problem, in America at least, that these sort of concepts are practically unheard of at the lower level mathematics (200 and down)

I have met about 40 people that told me, they had no idea what they were getting into when they chose to be a math major, because the stuff you do in calc and dif eq, and in alg, is nothing like the methods we use in topology /analysis.

So many people get scared of when they see these for the first time...

We could be HEROS!

Answer 33

How do you proof using epsilon-delta that if \(\lim_{x\to c} f(x)=L\neq 0\) then \(\lim_{x\to c} \frac{1}{f(x)}=\frac{1}{L}\)

Answer 34

I think the sequence characterization immediately follows epsilon delta stuff in any analysis book :

\(\lim \limits_{x\to a} f(x) = l \iff \) \(\forall \) sequences \(\{a_n\}\),
\(a_n \to a \implies f(a_n) \to l\)

compact sets is another neat definition if you really want to get fancy :D

Answer 35

much better, same with continuity

preimage of open set is open

BAM!

Answer 36

I don't really think about Epsilon Delta personally as my understanding of limits.
The way I understand them is that you are removing the boundary of the domain, and removing the boundary of the codomain as a result. You are showing that when all of the domain is finally cut away, the only thing left in the codomain is the limit being approached.

Answer 37

there is no epsilon delta in the book we teach this class out of(for limits)

its only the sequential version. @ganeshie8

Answer 38

of course you have \(\epsilon-N\) now, but that is easier to work with.

Answer 39

I see that works for limits of sequences but dont we need epsilon delta for functions ?

Answer 40

nope

Answer 41

the definitions are equivalent....

Answer 42

Right, im talking about \(\epsilon -N\) versus \(\epsilon - \delta \)

Answer 43

there are no "deltas" in sequence, and you notice the sequential definition says nothing of epsilon or delta

Answer 44

right, I think of N as a little easier d lol

Answer 45

if i remember correctly N is for limits at infinity

Answer 46

limit of sequences are always infinite

Answer 47

you literally don't need a delta in an analysis book

Answer 48

So who wants to tackle continuity?

Answer 49

```
preimage of open set is open
```

Is this the definition of continuity ?

Answer 50

one of them

Answer 51

its the topology definition

Answer 52

ill do an example of the epsilon delta proof so people can see how they are related

A function \(f:D\rightarrow \mathbb{R}\) is continuous at \(x_0\) provided
\(\forall \epsilon > 0 \exists \delta >0\forall x\in D |x-x_0|<\delta \implies |f(x) - f(x_0)|<\epsilon\)

Let us look at the example, \[f:(0,\infty) \underset{x\mapsto \frac{1}{x}}{\longrightarrow} \mathbb{R}\]

The first thing I do is consider what the result would look like.

\(|\frac{1}{x}-\frac{1}{c}|=\frac{|x-c|}{xc}\). The real problem here is the \(x\) on the bottom. Since \(c,\epsilon\) come before \(\delta\) in our definition, it is ok do define \(\delta\) dependent on them, but since \(x\) comes after \(\delta\) we can't have \(\delta\) depending on \(x\). (\(\delta\) has been chosen before \(x\) even entered in the picture). So it would be nice if we could find a number \(y\) s.t. we have \(\frac{1}{x}<y\) because then we could say \(\frac{|x-c|}{xc}<y\frac{|x-c|}{c}\). Then we could replace \(|x-c|\) with \(\delta\), then shove it all under \(\epsilon\) and solve for \(\delta\)(remember we are trying to hunt for a delta). So we would have \(y\frac{|x-c|}{c}<\epsilon\) then \(y\frac{\delta}{c}<\epsilon\) and we solve for \(\delta\) and get \(\delta < c\frac{\epsilon}{y}\) so we would have our delta, and we would be done. So now we find the \(y\). \\

One thing I always do is start with \(|x-c|<\delta\) and see what we can do. Well this implies \(c-\delta<x<c+\delta\) and if \(c-\delta>0\) we can do this \(\frac{1}{x}<\frac{1}{c-\delta}\), So we found our \(y\). To make sure \(c-\delta > 0\) we replace \(\delta\) with \(\frac{c}{2}\). So if \(\delta < \frac{c}{2}\) we have \(|x-c|<\delta \implies |x-c|<\frac{c}{2}\implies \frac{1}{x}< \frac{1}{c-\frac{c}{2}} = \frac{2}{c}\). \\

So now we have \(\frac{|x-c|}{xc}<\frac{2}{c}\frac{|x-c|}{c}=\frac{2|x-c|}{c^2}\) and now I replace \(|x-c|\) with \(\delta\) and shove it all under \(\epsilon\) and then solve for \(\delta\). \(\frac{2\delta}{c^2}<\epsilon \iff \delta < \frac{c^2\epsilon }{2}\). So now we tie it all together. \\ \\

proof:
Let \(c\in (0,\infty)\). Let \(\epsilon > 0\) be given. Set \(\delta < \min\{\frac{c}{2}, \frac{c^2\epsilon}{2}\}\). Then \(\forall x\in (0, \infty)\) we have, \[|x-c|<\delta \implies | \frac{1}{x}-\frac{1}{c} | = \frac{|x-c|}{xc} < \frac{2|x-c|}{c^2} < \frac{2\delta}{c^2}<\epsilon\] as required.

Answer 53

I wrote this for my class, then I put this example on the final lol

Answer 54

this shows its continuous for all x_0 in D=R^+

Answer 55

g.n. from canada

Answer 56

@watchmath The limit you gave tells me that:\[
0\lt |x-c|\lt \delta_1 \implies |f(x)-L|\lt \epsilon_1
\]There is a function \(\delta_1= g_1(\epsilon_1)\).\[
0\lt |x-c|\lt \delta \implies \left|\frac{1}{f(x)}-\frac 1L\right|\lt \epsilon\\
0\lt |x-c|\lt \delta \implies \left|\frac{L-f(x)}{f(x)L}\right|\lt \epsilon\\
0\lt |x-c|\lt \delta \implies \left|\frac{f(x)-L}{f(x)L}\right|\lt \epsilon\\
0\lt |x-c|\lt \delta \implies \left|f(x)-L\right|\lt \epsilon|f(x)L|
\]We want to say: \[
\delta = g_1(\epsilon|f(x)L|)=g_1(\epsilon_1)
\]
We know that \(L\neq 0 \implies 0\lt |L|\)
We know there is some \(\delta_0=g_1(\epsilon_0 )= g_1(|L|/2)\) such that:\[
|f(x)-L|<\frac{|L|}{2} \implies L-\frac{|L|}{2}<f(x)< L+\frac{|L|}{2}
\]If \(L\) is positive then:\[
L-\frac{|L|}{2} = \frac{L}{2} <f(x) \implies |L|/2 < |f(x)| \implies b= |L|^2/2 < |f(x)||L|
\]If it \(L\) is negative:\[
-\left(L+\frac{|L|}{2}\right) < -f(x) \implies -L/2 < -f(x) \implies b= |L|^2/2 < |f(x)||L|
\]So we let \[
\delta = \min\left(\delta_0= g_1\left(\frac{|L|}{2}\right), g_1\left(\frac{\epsilon|L|^2}{2}\right)\right)
\]I'm sorry it took so long. This thread got so laggy due to other people complaining in the thread. Also, I'm tired.

Answer 57

@watchmath The last part is really tricky, but you know how you said \(L\neq 0\)?

Well we know there is a \(\delta\) range that corresponds to an \(\epsilon\) range which will ensure that \(f(x)\) closer to \(L\) than it is to \(0\). This means we can ensure that \(|L/2| <|f(x)|\). Then all we have to do is substitute in \(|L/2|\) for \(|f(x)|\) since we know it will be smaller.

Answer 58

One way to get good at these proofs is to prove certain limit rules. \[
\lim_{x\to a}c = c
\]So: \[
|x-a|<\delta \implies |c-c| < \epsilon \\
|x-a|<\delta \implies 0< \epsilon
\]There is no way for the right side to be false, which means the conditional is always true.

Answer 59

\[
\lim_{x\to a}x = a
\]So \[
|x-a|<\delta \implies |x-a|<\epsilon
\]Just let \(\delta = \epsilon\).

Answer 60

Thanks. What I do below is basically the same as what you did. But I think it is better for us to think how we can bound
\[\frac{1}{|f(x)|}\]
For \(x\) close enough to \(c\) say \(0<|x-c|<\delta_1\) one can make \(|f(x)-L|<|L|/2\).
But \(|L|-|f(x)|\leq |f(x)-L|<\frac{|L|}{2}\). Hence \(|f(x)|>|L|/2\) and
\[\frac{1}{|f(x)|}<2/|L|\]
Now there exist \(\delta_2>0\) so that for \(0<|x-c|<\delta_2\) one have \(|f(x)-L|<L^2\epsilon/2\).
Now if we take \(\delta:=\min\{\delta_1,\delta_2\}\) then for all \(0<|x-c|<\delta\) we have
\[\left |\frac{1}{f(x)}-\frac{1}{L}\right |=\frac{|f(x)-L|}{|L||f(x)|}<\frac{2|f(x)-L|}{L^2}<\epsilon\]

Answer 61

That's extremely close to what I did, but I tried to be extremely rigorous because it is easy to make a wrong assumption for these proofs.

Answer 62

One funny thing about these proofs if that we always do them backwards.
They say "don't assume what you are trying to prove is true", however you can start with what you're trying to prove so long as all your steps are bi-conditional.

Answer 63

If we were to write the proofs in order, they'd be even more confusing.

Answer 64

I feel like epsilon delta proofs just sort of tuck the infinities away and that it's really just throwing away calculus as it was discovered and creating a forgery that acts and appears like calculus, but really isn't. So what do you say to that? haha =P

Answer 65

I don't really see how that is the case

Answer 66

Epsilon-delta proof is just a way to ensure the rigorousness of intuition behind calculus. Surely the intuition is lost in this delta-epsilon yoga thingy. If you still want to do calculus while it is still intuitive (say the idea of infinitesimal) you may want do do calculus using "non standard" analysis.

Answer 67

I think limits are inherently more complicated than they would seem intuitively, and that the Epsilon-Delta gives a really good understanding of limits.

I think most people who have an intuitive understanding of limits but struggle with Epsilon -delta don't really understand limits as well on a technical level.

Answer 68

Also, the definition could be extended to any algebra with a metric operation between the inputs of a function and the outputs of a function.

Answer 69

@Kainui I am confused by this statement. This is the rigor of calculus. Also how we define it....

Answer 70

I feel calc without delta epsilon is not calc at all, but just playing with what calc provides.

Answer 71

How come there isn't a single epsilon or delta in the work of Newton and Leibniz?

Answer 72

The greeks invented geometry and didn't know about irrational numbers. what is your point?

Answer 73

What about the work by millions of people after Newton...

Answer 74

What about Hilbert.

Answer 75

lol

Answer 76

Look I'm not saying it's not nice to have things that are ambiguous cleared up, but it just feels like e/d proofs are more smoke in mirrors trying to explain "what" calculus is, but I don't believe it. I believe calculus, like all math, is a discovered phenomena and is just as scientific in its discovery as anything else. To say "we define it to work this way and set theory as its basis" just seems to be mathematicians fooling themselves into thinking math is invented and not discovered. Have you ever used the square root function and felt like the limitation that a function be 1-to-1 was sort of an artificial construction?

Answer 77

Its not smoke and mirrors its rigor. No offense but I wont even read the rest if you say its smoke and mirrors.

Of course Newton was more smoke in mirrors because it was not rigorus..

Answer 78

Forget I said anything.

Answer 79

Well if you refuse to read my arguments, then I'm gone.

Answer 80

I'm actually just disappointed. The worthwhile conversations are ones in which people disagree. It's a shame you are close minded about it.

Answer 81

hey @wio can you prove this for me using epsilon delta thingy
\[\lim_{x \rightarrow 0}\frac{1}{x^2}=\infty\]

Answer 82

\[|x-x_n|<\delta \implies |\frac{1}{x^2}-\frac{1}{x^2_n}|< \epsilon \\ \text{ } \\ \text{ } \text{ } \implies |\frac{x_n^2-x^2}{x^2x_n^2}|< \epsilon \implies |\frac{x^2-x^2_n}{x^2x_n^2}| < \epsilon \\ \text { } \text{ } \text{ } \implies |x-x_n| < \frac{\epsilon(x^2x_n^2)}{|x+x_n|}\]
would it go something like this for choosing delta
also x_n->

Answer 83

0

Answer 84

Sure. When it comes to limits that tend to infinity, you have to change the epsilon inequality (and when it comes to limits where the input tends to infinity, you have to change the delta inequality).
The definition for this case is: \[
\forall \epsilon \exists\delta\forall x\quad |x-0|<\delta\implies \epsilon<\frac{1}{x^2}
\]it simplifies to: \[
|x|<\delta\implies \epsilon<\frac{1}{x^2} \\
|x|<\delta\implies \epsilon<\frac{1}{|x|^2}\\
|x|<\delta\implies |x|^2<\frac{1}{\epsilon}\\
|x|<\delta\implies |x|<\sqrt{\frac{1}{\epsilon}}
\]So we can choose: \[
\delta = g(\epsilon) = \sqrt{\frac 1{\epsilon}}
\]

Answer 85

\[|x-0|<\delta=\sqrt{\frac{1}{\epsilon }} \\ \lim_{L \rightarrow \infty}|f(x)-L|=\lim_{x_n \rightarrow 0}|\frac{1}{x^2}-\frac{1}{x_n^2}| \]
\[\lim_{x_n \rightarrow 0}|\frac{x_n^2-x^2}{x^2x^2_n}| =\lim_{x_n \rightarrow 0}\frac{1}{x^2x_n^2} |x-x_n||x+x_n| <\lim_{x \rightarrow 0} \frac{1}{x^2x_n^2} \sqrt{\frac{1}{\epsilon } } |x+x_n|\]
I don't know if this is the correct way to contrust the proof after finding delta
and also I don't know how to finish it if it is

Answer 86

that last thing is x_n->0 (not x->0)

Answer 87

I'm not familiar with the exact definition you are using. If I were, I might be able to help more.

Answer 88

I was trying to show |f(x)-L|<epsilon for when |x-a|<delta

Answer 89

where L->infinity and a is 0

Answer 90

wait..
\[< \frac{1}{x^2\frac{1}{\epsilon}} \sqrt{\frac{1}{\epsilon}} \sqrt{\frac{1}{\epsilon }}=\frac{1}{x^2} \]?

Answer 91

so we aren't suppose to show <epsilon maybe

Answer 92

For limits headed towards infinity, we're not showing that the gap between them and infinity is shrinking... infinity will always be infinity away from a finite number.
Instead we show that we will always be able to go bigger any any number that is provided.

That is why we used \(\epsilon < f(x)\) instead of \(|f(x)-L|<\epsilon\).
In most cases they will even change \(\epsilon\) to \(N\) since \(\epsilon\) is associated with an extremely small positive value.

Answer 93

stupid question
would you consider what you to be the proof
or just the construction of delta
or both?

Answer 94

what you did*

Answer 95

What I did was essentially an attempt to find \(\delta = g(\epsilon)\).
Once I found \(g\), my proof would basically be writing all of my work backwards. I don't go through that labor because it is somewhat pointless.
The steps for these proofs tend to be bidirectional, so we don't have the issues that come with "assuming what is to be proven".

Answer 96

Also thanks for this tutorial
I'm really glad you brought this up so I could remember to bring up that question.

Answer 97

and if I wanted to prove this \[\lim_{x \rightarrow 0}\frac{1}{x^3} dne \]
i would do a proof by contradiction
assume the limit does exist and it is L
right?