# Proof of limits

1. Jun 30, 2015

### 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.

2. Jun 30, 2015

### 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 $\lim_{x\to\infty} \frac{1}{x} = 0$?

3. Jun 30, 2015

### Josh S Thompson

My professor told me the exact opposite about infinity.

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

4. Jun 30, 2015

### 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

5. Jun 30, 2015

### PeroK

"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.

6. Jun 30, 2015

### Josh S Thompson

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

7. Jun 30, 2015

### pwsnafu

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.

8. Jun 30, 2015

### 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.

Last edited by a moderator: Jun 30, 2015
9. Jun 30, 2015

### 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.

10. Jun 30, 2015

### Josh S Thompson

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.

11. Jun 30, 2015

### jbriggs444

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?

12. Jun 30, 2015

### 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

13. Jun 30, 2015

### jbriggs444

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.

14. Jun 30, 2015

### Staff: Mentor

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

15. Jun 30, 2015

### jbriggs444

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

16. Jun 30, 2015

### Staff: Mentor

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." ?

17. Jun 30, 2015

### Josh S Thompson

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.

18. Jun 30, 2015

### jbriggs444

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".

19. Jun 30, 2015

### Staff: Mentor

That's a very pedantic distinction.

20. Jun 30, 2015

### jbriggs444

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.
Code (Text):

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.

Last edited: Jun 30, 2015
21. Jun 30, 2015

### jbriggs444

This is mathematics.

22. Jun 30, 2015

### Staff: Mentor

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.

23. Jun 30, 2015

### jbriggs444

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.

Last edited: Jun 30, 2015
24. Jun 30, 2015

### micromass

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.

25. Jun 30, 2015

### Edgardo

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?"