Question about limits involving 1/x

  • Thread starter docnet
  • Start date
  • #1
docnet
587
242
Homework Statement:
##\frac{1}{x}##
Relevant Equations:
##\frac{1}{x}##
For values of ##x## such that ##x>0##, is ##\frac{1}{x}## bound above?

My reaction is that if ##x>0## then ##\frac{1}{x}## is defined because ##x\neq 0##. But, it is not bound above because ##x## can be taken arbitrarily close to ##0## and ##\displaystyle{\lim_{x \to 0}} \frac{1}{x} = \infty##
 

Answers and Replies

  • #2
fresh_42
Mentor
Insights Author
2022 Award
17,655
18,375
Homework Statement:: ##\frac{1}{x}##
Relevant Equations:: ##\frac{1}{x}##

For values of ##x## such that ##x>0##, is ##\frac{1}{x}## bound above?

My reaction is that if ##x>0## then ##\frac{1}{x}## is defined because ##x\neq 0##. But, it is not bound above because ##x## can be taken arbitrarily close to ##0## and ## \[ \lim_{x \to 0} \frac{1}{x} = \infty \]##
Yes. And what is the question?

You can formally say: Given ##M > 0,## then for all ##0< x < 1/M## we get ##\dfrac{1}{x}>M##. This means that we can find a number ##1/x## which is greater than any boundary ##M.##
 
  • Like
Likes nuuskur, docnet and MatinSAR
  • #3
Prof B
66
37
Homework Statement:: ##\frac{1}{x}##
Relevant Equations:: ##\frac{1}{x}##

For values of ##x## such that ##x>0##, is ##\frac{1}{x}## bound above?

My reaction is that if ##x>0## then ##\frac{1}{x}## is defined because ##x\neq 0##. But, it is not bound above because ##x## can be taken arbitrarily close to ##0## and ##\displaystyle{\lim_{x \to 0}} \frac{1}{x} = \infty##
##\displaystyle{\lim_{x \to 0}} \frac{1}{x}## is undefined
##\displaystyle{\lim_{x \to 0^+}} \frac{1}{x} = \infty##
Follow your teacher's instructions, but for questions like this I think you are better off using the actual definition of unbounded.
 
  • #4
docnet
587
242
Yes. And what is the question?

You can formally say: Given ##M > 0,## then for all ##0< x < 1/M## we get ##\dfrac{1}{x}>M##. This means that we can find a number ##1/x## which is greater than any boundary ##M.##
Kind of related to limits, how does one numerically/symbolicaly write "the smallest element greater than x"? Since the real numbers exist in a continuity, the intuition seems like there should be one single irrational number that is "touching x" like in the case of the integers, "x+1 is touching x"
 
  • #5
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,790
15,402
Kind of related to limits, how does one numerically/symbolicaly write "the smallest element greater than x"? Since the real numbers exist in a continuity, the intuition seems like there should be one single irrational number that is "touching x" like in the case of the integers, "x+1 is touching x"
There is no such number. The set ##\{y \in \mathbb R: y > x \}## has no least member.

In other words, that is a set without a minimum. The number ##x##, however, is the infimum of the set. The infimum is also know as the greatest lower bound. Compare with supremum, which is the least upper bound.

These are absolutely critical concepts in standard real analysis.
 
  • Like
Likes Mark44 and docnet
  • #6
docnet
587
242
There is no such number. The set ##\{y \in \mathbb R: y > x \}## has no least member.

These are absolutely critical concepts in standard real analysis.
I understand why.. given a ##y>x##, you can always write a smaller number ##\frac{y-x}{2}+x > x##. This has been bothering me because math seems to be broken somehow. It's weird that a set of numbers has a lower bound, but not a minimum element.
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,790
15,402
I understand why.. given a ##y>x##, you can always write a smaller number ##\frac{y-x}{2}+x > x##. This has been bothering me because math seems to be broken somehow. It's weird that a set of numbers has a lower bound, but not a minimum element.
Even the rationals have that property. There is no least rational greater than ##0##, for example.

That one's easy, because ##0 < \frac m {2n} < \frac m n##.
 
  • #8
fresh_42
Mentor
Insights Author
2022 Award
17,655
18,375
There is no such number. The set ##\{y \in \mathbb R: y > x \}## has no least member.

In other words, that is a set without a minimum. The number ##x##, however, is the infimum of the set. The infimum is also know as the greatest lower bound. Compare with supremum, which is the least upper bound.

These are absolutely critical concepts in standard real analysis.
... and there is the axiom of choice. I even bought a book because it contains the equivalence proofs of the various versions. And it turned out to be a highly technical analysis book (Hewitt, Stromberg).
 
  • #9
36,716
8,717
This has been bothering me because math seems to be broken somehow. It's weird that a set of numbers has a lower bound, but not a minimum element.
IMO, it's not math that's broken, but rather your concept of it is flawed. What you're describing is the difference between discrete sets such as the integers, which have uniform separations between adjacent members, versus sets such as the real numbers.
 
  • #10
nuuskur
Science Advisor
798
667
This has been bothering me because math seems to be broken somehow. It's weird that a set of numbers has a lower bound, but not a minimum element.
I might find some things in math strange, but it's not sufficient to blame math when it doesn't agree with intuition. ##\mathbb R## is not well ordered with respect to the natural order.
 
  • #11
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
It's weird that a set of numbers has a lower bound, but not a minimum element.
Yes it is kind of weird, but that's why one of the axioms on the set of real numbers is that every bounded set has its infimum and supremum, but not necessarily minimum or maximum. So minimum or maximum are the infimum (or supremum) that also belong to the set.

For example if we consider the set $$Y(x)=\{y \in \mathbb{R}: y>x\}$$ it is infimum of Y(x)=x but the minimum doesn't exist because ##x## doesn't belong in ##Y(x)## as x>x doesn't hold. If we change its definition to $$Y(x)=\{y \in \mathbb{R}: y\geq x\}$$ then the infimum and minimum exist and they are equal to ##x##.

And no, it doesn't matter if the Y set is countable or uncountable, or x is rational or irrational. To see this consider the set $$Y=\{\frac{1}{n} ,n\in \mathbb{N}\}$$. The set Y is countable, n is integer, 1/n is rational but it doesn't have minimum either, the infimum is 0, but there is no ##n\in\mathbb{N}## for which $$\frac{1}{n}=0$$.
 
Last edited:

Suggested for: Question about limits involving 1/x

Replies
1
Views
335
Replies
4
Views
337
Replies
7
Views
334
  • Last Post
Replies
10
Views
356
Replies
6
Views
678
  • Last Post
Replies
2
Views
261
Replies
10
Views
562
Replies
25
Views
238
Replies
1
Views
247
Top