Proof of limits

[Josh S Thompson]

How do you prove

1/x = 0 if x= infinite
and
1/x = infinite if x=0

No need to use formal language, please just explain what the variables are for the quantifiers, or what you mean by for all and there exists if you are gracious enough.

 

[pwsnafu]

First, if we are talking about the real numbers then they are not true, hence you cannot prove them.
Second, don't use "infinite" as a noun. It's "infinity".
Third, $\frac{1}{ \infty} = 0$ on the extended real numbers by definition. There is nothing to prove.
Fourth, $\frac{1}{0}$ is undefined on both the real numbers and extended real numbers.

Edit: Wait are you asking for a proof that $\lim_{x\to\infty} \frac{1}{x} = 0$?

 

[Josh S Thompson]

First, if we are talking about the real numbers then they are not true, hence you cannot prove them.
Second, don't use "infinite" as a noun. It's "infinity".
Third, $\frac{1}{ \infty} = 0$ on the extended real numbers by definition. There is nothing to prove.
Fourth, $\frac{1}{0}$ is undefined on both the real numbers and extended real numbers.

Edit: Wait are you asking for a proof that $\lim_{x\to\infty} \frac{1}{x} = 0$?

My professor told me the exact opposite about infinity.

What do you mean extended real numbers and what do you mean by definition

 

[Josh S Thompson]

Actually I think I used infinity as an adjective, and my professor stopped me and was like infinite and he yelled at me.

How can infinity be a noun if it means increases without bound

 

[PeroK]

Actually I think I used infinity as an adjective, and my professor stopped me and was like infinite and he yelled at me.

How can infinity be a noun if it means increases without bound

"Infinity" is a noun, but (like many nouns) can be use as an "attributive" noun, where it acts like an adjective. E.g. an infinity simulator.

"Infinite" is an adjective, but (like many adjectives) can be used as an "adjectival" noun. E.g. we are discussing the infinite here.

 

[Josh S Thompson]

First, if we are talking about the real numbers then they are not true, hence you cannot prove them.
Second, don't use "infinite" as a noun. It's "infinity".
Third, $\frac{1}{ \infty} = 0$ on the extended real numbers by definition. There is nothing to prove.
Fourth, $\frac{1}{0}$ is undefined on both the real numbers and extended real numbers.

Edit: Wait are you asking for a proof that $\lim_{x\to\infty} \frac{1}{x} = 0$?

These limits do not exist because you can always pick a higher number , y, such that 1/x > 1/y >0.
and
and you can pick a smaller number y such that infinity > 1/y > 1/x

 

[pwsnafu]

How can infinity be a noun if it means increases without bound

Nouns can represent concepts. "Infinity" means the concept of not being finite.

These limits do not exist because you can always pick a higher number , y, such that 1/x > 1/y >0.
and
and you can pick a smaller number y such that infinity > 1/y > 1/x

If we are talking about real-valued functions, then $\lim_{x\to0} \frac{1}{x}$ does not exist. However, $\lim_{x\to\infty}\frac{1}{x}$ does exist and is equal to zero.

"Infinite" is an adjective, but (like many adjectives) can be used as an "adjectival" noun. E.g. we are discussing the infinite here.

I've never seen infinite used as a noun in mathematics discourse. Philosophy yes, but never in mathematics.

 

[HallsofIvy]

Since what you say your professor did was outrageous and he is not here to defend himself, I am going to ignore that. I suspect you are misunderstanding.

To talk about "1/0" and "1/infinity" as if they were numbers is simply incorrect. But I notice that this thread is titled "Proof of limits". That's a completely different matter! The "limit as x goes to 0" or the "limit as x goes to infinity" is not saying "x is 0" or "x is infinity". "Going to infinity" is simply "shorthand" for "getting larger and larger without bound.

Since you talk about limits, I assume that you have seen the definition: We say that "the limit of f(x), as x goes to a, is L" or $\lim_{x\to a} f(x)= L$, if and only if "given any $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- L|< \epsilon$. Roughly speaking that means that we can get f(x) as close to L as we please by making x close enough to a.

For the symbol "infinity", which is not really a number, we interpret it as "getting larger and larger" $\lim_{x\to \infty} f(x)= L$ is defined as "given any $\epsilon> 0$ there exist X such that if x> X then $|f(x)- L|< \epsilon$. For $\lim_{x\to a} f(x)= \infty$ we have "given any M, there exist $\delta> 0$ such that if $|x- a|< \delta$ then f(x)> M.

To prove that $\lim_{x\to \infty} \frac{1}{x}= 0$, I would note that $|f(x)- L|= |\frac{1}{x}- 0|= \frac{1}{x}$ since x going to infinity means x must be positive. So I want to get $\frac{1}{x}< \epsilon$ which is the same as $\frac{1}{\epsilon}< x$. For any $\epsilon> 0$, $\frac{1}{\epsilon}$ exists. Take M in the definition to be $\frac{1}{\epsilon}$.

To prove $\lim_{x\to 0} \frac{1}{x}= \infty$, I would note that $f(x)= \frac{1}{x}> M$ is the same as $x> \frac{1}{M}$. So, given such an M, I would take $\delta= \frac{1}{M}$. Then if $|x- 0|< \delta = \frac{1}{M}$, $\frac{1}{x}> M$.

Notice that there is no use of a "number" infinity in either of those.

 

[Josh S Thompson]

What does for all epsilon>0 there exists delta > 0 mean?

Is the for all epsilon just a qualifier for delta, like you are defining what delta can be?


For the symbol "infinity", which is not really a number, we interpret it as "getting larger and larger" lim x→∞ f(x)=L \lim_{x\to \infty} f(x)= L is defined as "given any ϵ>0 \epsilon> 0 there exist X such that if x> X then |f(x)−L|<ϵ

How do you go from |x-a| < delta to x>X

Thank you very much I always wanted to understand this stuff.

 

[Josh S Thompson]

Nouns can represent concepts. "Infinity" means the concept of not being finite.



If we are talking about real-valued functions, then $\lim_{x\to0} \frac{1}{x}$ does not exist. However, $\lim_{x\to\infty}\frac{1}{x}$ does exist and is equal to zero.


I've never seen infinite used as a noun in mathematics discourse. Philosophy yes, but never in mathematics.

Idk man i confused not being real with not being a noun.


I just don't know what a limit is, you treat it like x=X but x does not equal X, that's probably what I mean when I said the limits do not exist.


Yea bro that was my philosophy teacher who said that I take philosophy classes because they are easier than math classes, everyone who gets a question wrong in math should switch majors to philosophy because its easier.

 

[jbriggs444]

What does for all epsilon>0 there exists delta > 0 mean?

Let's start with something easier. "for all epsilon > 0, epsilon² > 0" Can you figure out what that means? Is it a meaningful statement? Is it true or is it false?

 

[Josh S Thompson]

if all numbers in this problem are positive, the square of all numbers in the problem will be positive. that's how I interpret that statement

for all epsilon > 0, there exists delta > 0. ---- How is that meaningful, it says all the numbers in the problem are positive and there is a positive number that we call delta, who cares about for all. Please explain

 

[jbriggs444]

if all numbers in this problem are positive, the square of all numbers in the problem will be positive. that's how I interpret that statement

That is not correct. The statement "for all epsilon > 0, epsilon² > 0" makes no mention of all the numbers in "the problem". Nor does it take the form of an if then statement.

It is an assertion that every number that is greater than zero has a square that is greater than zero.

 

[Mark44]

That is not correct. The statement "for all epsilon > 0, epsilon² > 0" makes no mention of all the numbers in "the problem". Nor does it take the form of an if then statement.

"If - then" is implied. One could read the above as saying, "if $\epsilon$ is any positive number, then $\epsilon^2$ is positive."

 

[jbriggs444]

"If - then" is implied. One could read the above as saying, "if $\epsilon$ is any positive number, then $\epsilon^2$ is positive."

That is not correct. The for all is outside the if. You are simply mistaken.

 

[Mark44]

That is not correct. The for all is outside the if. You are simply mistaken.

You're going to have to do better than that.

How does "for all epsilon > 0, epsilon² > 0" differ in meaning from "if $\epsilon$ is any positive number, then $\epsilon^2$ is positive." ?

 

[Josh S Thompson]

That is not correct. The statement "for all epsilon > 0, epsilon² > 0" makes no mention of all the numbers in "the problem". Nor does it take the form of an if then statement.

It is an assertion that every number that is greater than zero has a square that is greater than zero.

Why can't you just say; x^2 > 0, where x > 0.

So for all epsilon > 0, there exists delta > 0 means for every number greater than 0 there exists a delta greater than 0. but that doesn't make sense you need to define that relationship further. and if you define the relationship further like in the limit definition, you need to solve the proof for the statement to make sense.

 

[jbriggs444]

You're going to have to do better than that.

How does "for all epsilon > 0, epsilon² > 0" differ in meaning from "if $\epsilon$ is any positive number, then $\epsilon^2$ is positive." ?

The one is a formula with no free variables. The other is a formula with one free variable.

Edit, I should be more clear here.

It is common usage to interpret an if statement with one free variable as if it were implicitly qualified with a "for all" on the free variable. However, such a statement is then formally a "for all" rather than an "if then".

 

[Mark44]

The one is a formula with no free variables. The other is a formula with one free variable.

That's a very pedantic distinction.

 

[jbriggs444]

Why can't you just say; x^2 > 0, where x > 0.

Edit: Let me try to address this in more detail and in the process try to make it more clear why the distinction that Mark44 finds pedantic is one that I consider important.

Yes. You can just say "x^2 > 0 where x > 0". Or "if x > 0 then x^2 > 0". The reader will normally assume that you mean to make an assertion about all possible x, not just an assertion about some specific x.

To be pedantic, the logical formula: "if x > 0 then x^2 > 0" has one free variable, x. Its truth value could conceivably depend on the choice of x. By interpreting it as an assertion about all possible x, the reader is instead taking it as a shorthand way of expressing the logical formula: "for all x, if x > 0 then x^2 > 0". The latter formula has no free variables. It is either true (the if-then statement holds for all x) or it is false (the if-then statement fails to hold for at least one x).

The short-hand rule of interpretation is handy. The writer has less to write. The reader has less to read. Everybody wins.

But if one is nesting logical formulas inside of logical formulas, things can get difficult to interpret clearly. In the case of a definition of limits we have [at least] a "for all" enclosing a "there exists" which in turn encloses another "for all". In order to improve clarity, the "for all" statements are made explicitly. This way, the reader is not forced to assume anything to make sense out of the statement.

Here is a formula with indentation to indicate the nesting.

for all delta > 0
   there exists an epsilon > 0 such that
      for all x > epsilon
          |1/x - 0| < delta

So for all epsilon > 0, there exists delta > 0 means for every number greater than 0 there exists a delta greater than 0.[/QUOTE]
No. That is not correct. The statement goes on to make a further qualification on the relationship that some such delta has with the epsilon.

 

[jbriggs444]

That's a very pedantic distinction.

This is mathematics.

 

[Mark44]

You're going to have to do better than that.

How does "for all epsilon > 0, epsilon² > 0" differ in meaning from "if $\epsilon$ is any positive number, then $\epsilon^2$ is positive." ?

The one is a formula with no free variables. The other is a formula with one free variable.

Edit, I should be more clear here.

It is common usage to interpret an if statement with one free variable as if it were implicitly qualified with a "for all" on the free variable. However, such a statement is then formally a "for all" rather than an "if then".

First, such exceedingly fine points do nothing whatsoever to help the OP understand limits.
Second, your point might be of interest to a logician, but I would venture that most other mathematicians would call your point a difference without a distinction. Your formulation and mine convey exactly the same meaning. If you disagree, please show me a single value of $\epsilon$ for which one statement is true and the other is not.

 

[jbriggs444]

First, such exceedingly fine points do nothing whatsoever to help the OP understand limits.
Second, your point might be of interest to a logician, but I would venture that most other mathematicians would call your point a difference without a distinction. Your formulation and mine convey exactly the same meaning. If you disagree, please show me a single value of $\epsilon$ for which one statement is true and the other is not.

Epsilon is not a free variable for my formula, hence your challenge fails -- it is not a well formed question.

Edit: To be completely correct, I should retract that. A formula with no free variables can nonetheless continue to be true or false if the value of some extraneous variable has some particular value. I need to look to see what formula you have in mind in this case.

In this case, the statement you have chosen was if $\epsilon$ is any positive number, then $\epsilon^2$ is positive. And indeed that statement is both true as a universal statement and true of every epsilon.

If the statement had instead been if $\epsilon$ is any positive number then $\epsilon > 100$ then, when interpreted as a statement about all epsilon, that statement is false but when interpreted as a statement about a particular epsilon, it may be either true or false.

 

[micromass]

Epsilon is not a free variable for my formula, hence your challenge fails -- it is not a well formed question.

Edit: To be completely correct, I should retract that. A formula with no free variables can nonetheless continue to be true or false if the value of some extraneous variable has some particular value. I need to look to see what formula you have in mind in this case.

In this case, the statement you have chosen was if $\epsilon$ is any positive number, then $\epsilon^2$ is positive. And indeed that statement is both true as a universal statement and true of every epsilon.

If the statement had instead been if $\epsilon$ is any positive number then $\epsilon > 100$ then, when interpreted as a statement about all epsilon, that statement is false but when interpreted as a statement about a particular epsilon, it may be either true or false.

And I hope you are aware that many mathematics books and papers are written where the statement "If $\varepsilon>0$, then $\varepsilon^2>0$" is perfectly ok? You are not really helping the OP at this point, you're just confusing him.

 

[Edgardo]

What does for all epsilon>0 there exists delta > 0 mean?
Is the for all epsilon just a qualifier for delta, like you are defining what delta can be?

One way to understand this is to interpret it as a two-player game, let's say between you and me. The game works as follows:

I give you a number $\epsilon > 0$, and your task is to find a number $\delta > 0$ such that a certain condition is fulfilled.

You can read about this two-player game interpretation in this pdf document. If you want to find out more about the precise meaning of infinity then I recommend watching this youtube video, with the title "What does it mean to say that the limit of f(x) equals infinity?"

 

[Josh S Thompson]

I give you a number ϵ>0 \epsilon > 0, and your task is to find a number δ>0 \delta > 0 such that a certain condition is fulfilled.

I don't understand, epsilon is any and all numbers greater than 0. Why is this not defined as a set?

So I would be looking for a number, delta, that is also epsilon?

Does
there exists a number delta>0, for all epsilon >0, mean you have defined a specific delta>0 for every number greater than 0.
Is it different from
There exists a number delta>0, such that for all epsilon >0

To me, from my understanding of English the "such that" makes epsilon the subject and changes the whole sentence

 

[Josh S Thompson]

Thanks everyone for your comments you are helping but there is no need for pedantics, just the basic concepts, examples are best especially for this I read the definitions a 100 times but no examples with multiple qualifiers and paraphrases. Help is greatly appreciated.

 

[Mark44]

I don't understand, epsilon is any and all numbers greater than 0. Why is this not defined as a set?

No. $\epsilon$ is a particular (positive, and usually close to 0) number that someone tells you. Your part of the dialog is to find a positive number $\delta$ so that some condition is met.

If the other person is satisfied that you have established the limit, you're done. If not, the other person will make it harder, by telling you a smaller positive number, $\epsilon$. You then have to find a corresponding, but possibly different value of $\delta$. The process continues until the other person finally gives up.

So I would be looking for a number, delta, that is also epsilon?

Does
there exists a number delta>0, for all epsilon >0, mean you have defined a specific delta>0 for every number greater than 0.
Is it different from
There exists a number delta>0, such that for all epsilon >0

To me, from my understanding of English the "such that" makes epsilon the subject and changes the whole sentence

 

[Mark44]

An example might be helpful. Prove that $\lim_{x \to 2} 3x = 6$
This example is somewhat artificial, since the function involved is continuous everywhere, so you can simply substitute 2 into the function (f(x) = 3x) and see that the function value is 6. Nevertheless, the process I'll show can be applied to more complicated functions that you can't simply find the limit by substituting the value.

To prove the limit above, we need to show that for any $\epsilon$ > 0 that someone else chooses, we can find a $\delta$ > 0 so that, if |x - 2| < $\delta$, then |f(x) - 6| < $\epsilon$. That is, that |3x - 6| < $\epsilon$.

I note that |3x - 6| < $\epsilon$ is equivalent to 3|x - 2| < $\epsilon$, from which we get |x - 2| < $\frac{\epsilon} 3$

In this case, I can choose $\delta = \frac{\epsilon} 3$

So whatever the other person chooses for $\epsilon$, I have a formula for finding the $\delta$ that "works", that depends on the other person's choice of $\epsilon$. As long as x is within $\delta$ units of 2 (i.e., |x - 2| < $\delta$), then 3x will be within $\epsilon$ units if 6 (i.e., |3x - 6| < $\epsilon$).

 

[micromass]

Epsilon-delta definitions actually have a lot to do with basic physics. It's the same idea of an approximation.

Let's say that you're in a shower. You like your shower to have a temperature of 40°C or approximate. You can regulate your temperature by turning some knob. Of course, you can never get exactly 40°C by turning the knob (you don't have that precision), but you can get close. In fact you can get arbitrary close.

For example, let's say that you like 40°C, but ±5°C is ok too. Then you got a certain number of positions that are ok. In fact, you got an entire range of positions that is ok. The 5°C is called the ε, while the range of positions has to do with the δ.
However, if you're more sensitive and if you like 40°C, but only ±0.5°C is ok. Then there are also some positions of the knob that are ok, but there are significantly less positions that are ok.

In general, let's say that you like 40°C with a tolerance of ε°C. Then there is a certain range of positions of the knob that are ok. This is the δ-range. The smaller you take ε, the smaller the δ-range will be. But whatever we see, there will always be a δ-range.

So choosing your temperature in the shower is continuous.

An example of a discontinuous function would be the following. Consider the following graph and imagine that it is landscape: (the following graph IS continuous, but that's not the function I'm talking about)

http://t3.gstatic.com/images?q=tbn:ANd9GcTX168KV3vYwPP29L1p5ENwhcLZi-nBQ6FKdluQlJpHiLCtghLbGQ [Broken]

So your landscape is flat between 1 and 3, and decreases outside that. Let's say that you have a ball that you want to place on the landscape. Your goal is to place the ball exactly on 3. Of course, this is impossible to do. So we will allow some degree of tolerance. Let's say we want to place the ball 1 distance from the spot 3. We can always do this by placing it on the left of 3. But we can't do that by placing it on the right. Indeed, if we place it on the right, then the ball will just roll away and will roll outside of the allowable range. So we see that there is no allowable interval of tolerance around 3: the space left to 3 is ok, but the space right to 3 is not ok. This means that the process is not continuous.

Abstractly, you work with functions $f:\mathbb{R}\rightarrow \mathbb{R}$. You should see the domain as some kind of knob in the shower. So we can put the knob on any value we please , but only approximately. The function $f$ regulates the temperature. So let's say we put the knob on $0$, then we will feel temperature $f(0)$. The entire process of epsilon-delta definitions is to get as close as we want to some specific temperature by turning the knob and thus getting within a certain allowable range of the knob. If no matter how close we want to get to the temperature, there is always an allowable interval of the knob to the left and to the right, then the function is continuous.

 

[Josh S Thompson]

No. ϵ\epsilon is a particular (positive, and usually close to 0) number that someone tells you. Your part of the dialog is to find a positive number δ\delta so that some condition is met.

If the other person is satisfied that you have established the limit, you're done. If not, the other person will make it harder, by telling you a smaller positive number, ϵ\epsilon. You then have to find a corresponding, but possibly different value of δ\delta. The process continues until the other person finally gives up.

Wow, thank you that is a great explanation,
There are multiple turns in the game and it repeats at the beginning of the statement.

But what if there exists goes first, would you just have to prove the statement for one delta and all epsilon and the game would not repeat.

 

[Mark44]

Wow, thank you that is a great explanation,
There are multiple turns in the game and it repeats at the beginning of the statement.

But what if there exists goes first, would you just have to prove the statement for one delta and all epsilon and the game would not repeat.

That's not how it works. You can't arbitrarily change the definition of $\lim_{x \to a} f(x) = L$, which goes like this:
"For each $\epsilon > 0$ there exists a $\delta > 0$ such that if $0 < |x - a| < \delta, |f(x) - L| < \epsilon$."
Sometimes you see this with more symbolism, but this captures the idea of the definition.

You start with $\epsilon$ that someone gives you, and you find $\delta$ so that when x is within $\delta$ of a, then f(x) is within $\epsilon$ of L.

 

[Helolo]

$1/x$ doesn't eaqual to $\infty$ if $x=0$ and $1/x$ doesn't equal to $0$ if $x=\infty$, but the limit of 1/x when x approach ∞ or x approach to 0 it does and by the way Josh S. Thompson u need to learn about Limitations again. ☺

 

[Josh S Thompson]

If the other person is satisfied that you have established the limit, you're done. If not, the other person will make it harder, by telling you a smaller positive number, ϵ\epsilon. You then have to find a corresponding, but possibly different value of δ\delta. The process continues until the other person finally gives up.

What if the person picking epsilon never gives up?

 

[micromass]

What if the person picking epsilon never gives up?

Then the game goes on forever. The point is that the person choosing the $\delta$ can never lose.