# Inconsistencies in the concept of limits

• I

## Main Question or Discussion Point

I understand that 0.9999.... = 1 is true because in limit theory "getting arbitrarily close" means that they actually are equal. I'm also aware of the epsilon-delta definition of limits. But I feel like there are some inconsistencies in this concept; I'll explain using two scenarios.

Case 1: lim(X → 0) X = 0
When X gets arbitrarily close to zero... the function (which is actually just X as well) is equal to zero.

Case 2: lim(h → 0) [(X+h) - X] / h = h / h = 1
Here we argue that the derivative of the function X is 1. Because even though h is arbitrarily close to zero, it isn't equal to zero. So h / h is defined and is equal to 1.

***
This seems inconsistent; in order to keep in line with the concept that h "getting arbitrarily close" to zero means that h actually is equal to zero, shouldn't h / h really just be 0 / 0 and hence undefined?

Furthermore, it is also commonly know that in the topic of limits, 0 / 0 is considered one of the indeterminate forms, so why is this derivative even technically solvable?

## Answers and Replies

PeroK
Science Advisor
Homework Helper
Gold Member
I understand that 0.9999.... = 1 is true because in limit theory "getting arbitrarily close" means that they actually are equal. I'm also aware of the epsilon-delta definition of limits. But I feel like there are some inconsistencies in this concept; I'll explain using two scenarios.

Case 1: lim(X → 0) X = 0
When X gets arbitrarily close to zero... the function (which is actually just X as well) is equal to zero.

Case 2: lim(h → 0) [(X+h) - X] / h = h / h = 1
Here we argue that the derivative of the function X is 1. Because even though h is arbitrarily close to zero, it isn't equal to zero. So h / h is defined and is equal to 1.

***
This seems inconsistent; in order to keep in line with the concept that h "getting arbitrarily close" to zero means that h actually is equal to zero, shouldn't h / h really just be 0 / 0 and hence undefined?

Furthermore, it is also commonly know that in the topic of limits, 0 / 0 is considered one of the indeterminate forms, so why is this derivative even technically solvable?
It's an important question. But, you've missed the key point. If the terms in a sequence get arbitrarily close to a number then the limit of that sequence is equal to that number.

This is a subtle difference, but you need to distinguish between the individual terms of a sequence and the concept of the limit of a sequence.

In the derivative case, every term in the sequence is well-defined, as ##h## is never ##0##. The limit of that sequence is well-defined but it is not what you get by setting ##h = 0##.

Valour549
I don't get what this has to do with sequence. Maybe I can write out my logic below step-by-step so you guys can point out which part is flawed.

1) The statement, 0.9999..... = 1.0000.....1 = 1

I think it's true (because of this http://www.purplemath.com/modules/howcan1.htm)

2) When we write the given condition under the word lim , for example, limX→1, we mean that X = 0.9999..... or 1.0000.....1

3) Due to 1) above, X is also = 1

4) I now set a function f(x)=X, so the value of the function is actually identical to the variable in the condition, therefore I conclude limX→1 X = 1

5) However, if I chose a different function, and after simplification the limit looked like this: limX→1 (X-1)/(X-1), then I should conclude that the limit does not exist. Because rather than arguing that X-1 isn't zero, I should be consistent with 3) above and argue that X-1 is actually zero.

PeroK
Science Advisor
Homework Helper
Gold Member
5) However, if I chose a different function, and after simplification the limit looked like this: limX→1 (X-1)/(X-1), then I should conclude that the limit does not exist. Because rather than arguing that X-1 isn't zero, I should be consistent with 3) above and argue that X-1 is actually zero.
No, because in general:

##\lim (\frac{f(x)}{g(x)}) \ne \frac{\lim f(x)}{\lim g(x)}##

In particular, it's not true when ##\lim f(x) = \lim g(x) = 0##

Ah yes, because the limit is only taken after the function itself is first resolved, in that particular case.

Are there any problems with my parts 1) to 4) ?

To be honest I want to know what x→1 even means. Because if it means that x is arbitrarily close to 1, such that we cannot even differentiate between x and 1, then why doesn't it just say x=1 in the first place?

PeroK
Science Advisor
Homework Helper
Gold Member
Ah yes, because the limit is only taken after the function itself is first resolved, in that particular case.

Are there any problems with my parts 1) to 4)
You are not really saying anything in 1) to 4) that I can see.

Sorry see edit.

PeroK
Science Advisor
Homework Helper
Gold Member
Sorry see edit.
There's only one number ##1##. It does, however, have two decimal expansions: ##1.000 \dots ## and ##0.999 \dots##.

This is, in fact, true for all terminating decimals. E.g. ##2.5## can also be written as ##2.4999 \dots##.

So does x→1 mean x = 1.000... or 0.999... or 1, or none of these? If none of these, what exactly does x→1 mean beyond a vague lexical "...as X approaches 1"?

PeroK
Science Advisor
Homework Helper
Gold Member
So does x→1 mean x = 1.000... or 0.999... or 1, or none of these? If none of these, what exactly does x→1 mean beyond a vague lexical "...as X approaches 1"?
##x \rightarrow 1## is notation, used in the definition of a limit.

Outside of this, it can be thought of informally as "as x approaches 1". It certainly doesn't mean "let ##x = 1##"

You need to lose the idea that ##0.999 \dots## is anything other than an alternative way to write ##1##. It's nothing special!

Last edited:
To say that x→1 is simply a notation or that it can be thought of as as x approaches 1 is vague, because naturally one would ask "What do you mean by approach 1? How close is x to 1 really?"

But I think I thought of the answer to my own question...

Using the definition of a limit, "x is delta-close to 1, where the size of delta is as required by the chosen epsilon", but because the function may or may not be defined where x = 1, the size of the required delta is always greater than zero. Thus we know x→1 does not mean x = 1.

PeroK
Science Advisor
Homework Helper
Gold Member
To say that x→1 is simply a notation or that it can be thought of as as x approaches 1 is vague, because naturally one would ask "What do you mean by approach 1? How close is x to 1 really?"

But I think I thought of the answer to my own question...

Using the definition of a limit, "x is delta-close to 1, where the size of delta is as required by the chosen epsilon", but because the function may or may not be defined where x = 1, the size of the required delta is always greater than zero. Thus we know x→1 does not mean x = 1.
That's not a good way to think of it.

$$\lim_{x \rightarrow 1} f(x)$$

Only has a meaning in its entirety. ##\lim## and ##x \rightarrow 1## have no independent meaning in their own right. It's only when you put them together that is mathematically well-defined. And, in fact, you need a function ##f## with certain properties (defined in an open neighbourhood of ##1##) to be precise about what a limit is.

For example, you might as well ask: what does the "l" in "lim" mean?

Informally, you can talk about limits and x approaching 1. But, the formal mathematics requires both to be put together before you have something well defined. When you put this together, there is nothing vague.

Does the fact that we can write 0.999... (recurring) = 1 have anything to do with the concept of limits?

PeroK
Science Advisor
Homework Helper
Gold Member
Does the fact that we can write 0.999... (recurring) = 1 have anything to do with the concept of limits?
Yes. Before you define limits ##0.999 \dots## is meaningless. You need some way to interpret this. The most obvious interpretation is an "infinite sum":

##0.999 \dots = 0.9 + 0.09 + 0.009 + \dots##

Now, somewhat informally, this is covered in high school, as this is an infinite geometric series, with first term ##a = 0.9## and common ratio ##r = 0.1##. And, from high school, you know that:

##a + ar + ar^2 + ar^3 + \dots = \frac{a}{1-r} \ \ (|r| < 1)##

Hence:

##0.9 + 0.09 + 0.009 + \dots = \frac{0.9}{1- 0.1} = 1##

When you define and study limits, you can put infinite series, calculus and a whole lot else on a firm, rigorous footing. That's where the formal definition of a limit comes in.

Gotcha... thanks!

Mark44
Mentor
I don't get what this has to do with sequence. Maybe I can write out my logic below step-by-step so you guys can point out which part is flawed.

1) The statement, 0.9999..... = 1.0000.....1 = 1
The expression in the middle, above, is meaningless, for the reason that you can't specify where that final 1 is.
0.9999... is meaningful, however, as the ellipsis (...) denotes that every digit after the decimal point is a 9.

vanhees71
Science Advisor
Gold Member
2019 Award
It's very simple to answer this question, when remembering that the real numbers are definable as an equivalence class of ##\mathbb{Q}##-valued Cauchy sequences. I.e., you define all Cauchy series with the same limit as representing the same real number.

This immediately leads to the formal resolution of this problem. You can just use the ovious convergent geometric series,
$$0.9999...=\sum_{k=1}^{\infty} \frac{9}{10^k}=9 \frac{1/10}{1-1/10}=\frac{9}{10} \frac{1}{\frac{9}{10}}=1.$$

Stephen Tashi
Science Advisor
To be honest I want to know what x→1 even means.
The "liberal arts" approach to analyzing vocabulary and notation in mathematics isn't reliable.

In the liberal arts, people are trained to analyze statements by breaking them down in to words and phrases and analzying the meaning of these individual pieces. Mathematical definitions cannot be reconstructed by such an analysis. In the notation "##\lim_{x\rightarrow a} f(x) = L## there is no separate interpretation for the symbols "##x \rightarrow a##". Likewise in the verbal interpretation of that notation as "The limit of the function f(x) as x approaches a is equal to L", the definition of limit (the epsilon-delta definition) does not give a definition for the word "approaches" or the phrase "x approaches a".

You can think informally of the intuitive concept of limit being made up of smaller concepts and formulate a private and individualized picture of how limits work. Everyone does this, but this sort of thinking doesn't correspond to formal mathematics.

The structure of a mathematical definition can be put in the form:
[Something using some undefined notation or words] means [a statement using only previously defined notation and words].

The definition of "##\lim_{x\rightarrow a} f(x) = L##" defines it as a statement that begins "For each ##\epsilon > 0##, there exists a ##\delta > 0## such that....'. The definition does not define individual parts such as "##\lim##", "##x \rightarrow a##", or "##(x)##".