Otdfbt

What's currently blocking the construction of the wall between Mexico and the US?

How do I turn off a repeating trade?

How long would it take to cross the Channel in 1890's?

Do I need a shock-proof watch for cycling?

Dystopia where people are regularly forced to move, protagonist must move daily

Is a single radon-daughter atom in air a solid?

What exactly is the 'online' in OLAP and OLTP?

Parameterize chained calls to a utility program in Bash

Is this proposal by U.S. presidential candidate Pete Buttigieg to change the composition of the Supreme Court constitutional?

Can there be an UN resolution to remove a country from the UNSC?

What reason would an alien civilization have for building a Dyson Sphere (or Swarm) if cheap Nuclear fusion is available?

Why tighten down in a criss-cross pattern?

What does it mean to "control target player"?

Why do even high-end cameras often still include normal (non-cross-type) AF sensors?

How much will studying magic in an academy cost?

Can White Castle?

What did River say when she woke from her proto-comatose state?

Is it damaging to turn off a small fridge for two days every week?

Can Ogre clerics use Purify Food and Drink on humanoid characters?

How do I professionally let my manager know I'll quit over smoking in the office?

What size of powerbank will I need to power a phone and DSLR for 2 weeks?

How does DC work with natural 20?

How to make clear to people I don't want to answer their "Where are you from?" question?

Why do textbooks often include the solutions to odd or even numbered problems but not both?

What is the difference between “arbitrarily close” and “sufficiently close” in term of limits?

Understanding limits and how to interpret the meaning of “arbitrarily close”What is the difference between writing $f$ and $f(x)$?What is the significance of limit and why is there always a difference between the value of x and the limiting value?Understanding of the formal and intuitive definition of a limitthe need for the formal definition of a limitUnderstanding limits and how to interpret the meaning of “arbitrarily close”Why can we change a limit's function/expression and claim that the limits are identical?What is the difference between Cauchy's Rule and L'Hôpital's Rule?equality of a series to its limit at infinityWhat is it to be definedExponential gets arbitrarily close to a constant in the limit

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

$begingroup$

The definition of limit as always

“the limit of $f(x)$, as $x$ approaches $a$, equals $L$” means we can
make the values of $f(x)$ arbitrarily close to $L$ by restricting x to
be sufficiently close to $a$ but not equal to $a$.

What exactly mean by phrases "arbitrarily close" for $f(x)$ and "sufficiently close" for $x$ ? Are they interchangeable ?

calculus limits terminology

share|cite|improve this question

edited Jun 5 at 9:26

user1551

76.4k668130

asked Jun 5 at 9:23

Ammar BamhdiAmmar Bamhdi

758

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Teh closedness in sufficiently close depends on the one in arbitrarily close.
  $endgroup$
  – Kavi Rama Murthy
  Jun 5 at 9:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/a/1733170/72031
  $endgroup$
  – Paramanand Singh
  Jun 5 at 16:36
  

  
  
  
  

add a comment | 

3

$begingroup$

The definition of limit as always

“the limit of $f(x)$, as $x$ approaches $a$, equals $L$” means we can
make the values of $f(x)$ arbitrarily close to $L$ by restricting x to
be sufficiently close to $a$ but not equal to $a$.

What exactly mean by phrases "arbitrarily close" for $f(x)$ and "sufficiently close" for $x$ ? Are they interchangeable ?

calculus limits terminology

share|cite|improve this question

edited Jun 5 at 9:26

user1551

76.4k668130

asked Jun 5 at 9:23

Ammar BamhdiAmmar Bamhdi

758

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Teh closedness in sufficiently close depends on the one in arbitrarily close.
  $endgroup$
  – Kavi Rama Murthy
  Jun 5 at 9:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/a/1733170/72031
  $endgroup$
  – Paramanand Singh
  Jun 5 at 16:36
  

  
  
  
  

add a comment | 

$begingroup$

The definition of limit as always

“the limit of $f(x)$, as $x$ approaches $a$, equals $L$” means we can
make the values of $f(x)$ arbitrarily close to $L$ by restricting x to
be sufficiently close to $a$ but not equal to $a$.

What exactly mean by phrases "arbitrarily close" for $f(x)$ and "sufficiently close" for $x$ ? Are they interchangeable ?

calculus limits terminology

share|cite|improve this question

edited Jun 5 at 9:26

user1551

76.4k668130

asked Jun 5 at 9:23

Ammar BamhdiAmmar Bamhdi

758

$endgroup$

share|cite|improve this question

edited Jun 5 at 9:26

user1551

76.4k668130

asked Jun 5 at 9:23

Ammar BamhdiAmmar Bamhdi

758

share|cite|improve this question

edited Jun 5 at 9:26

user1551

76.4k668130

asked Jun 5 at 9:23

Ammar BamhdiAmmar Bamhdi

758

edited Jun 5 at 9:26

user1551

76.4k668130

edited Jun 5 at 9:26

user1551

76.4k668130

asked Jun 5 at 9:23

Ammar BamhdiAmmar Bamhdi

758

asked Jun 5 at 9:23

Ammar BamhdiAmmar Bamhdi

758

3 Answers
3

active

oldest

votes

5

$begingroup$

the limit of $f(x)$, as $x$ approaches $a$, equals $L$” means we can
make the values of $f(x)$ arbitrarily close to $L$

Rephrasing: "as close as we want"

by restricting x to be sufficiently close to $a$ but not equal to $a$.

Rephrasing: "close enough"

---

Or put differently: $L$ is the limit (as $x$ approaches $a$) if we can

• make the distance from $f(x)$ to $L$ as small as we want,


• by only making the distance from $x$ to $a$ small enough.

share|cite|improve this answer

edited Jun 6 at 7:40

answered Jun 5 at 9:27

StackTDStackTD

25.3k2254

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there something wrong with this answer? If the down voter cares to explain, perhaps I can improve it.
  $endgroup$
  – StackTD
  Jun 5 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't downvote, but I suggest replacing "only by" (suggesting "this is the only way") with "by only" (suggesting "using this as our only tool to achieve the desired effect, even though there could be other ways").
  $endgroup$
  – Aasmund Eldhuset
  Jun 5 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AasmundEldhuset Thanks, that's better indeed.
  $endgroup$
  – StackTD
  Jun 6 at 7:40
  

  
  
  
  

add a comment | 

2

$begingroup$

• arbitrarily close: true for any closeness, how large or how small; $forallepsilon$.

$|x-1|>0$ holds for $x$ arbitrarily close to $1$.

• sufficiently close: true under some suitable upper bound; $forallepsilon<u$ where $u$ is dictated by the problem on hand.

$|x-1|<1$ holds for $x$ sufficiently close to $1$.

share|cite|improve this answer

edited Jun 5 at 12:46

answered Jun 5 at 9:40

Yves DaoustYves Daoust

139k880237

$endgroup$

add a comment | 

1

$begingroup$

You do not mention where your function $f$ is defined and which range it has, but let us assume that it is a function $f : mathbb R to mathbb R$. Your definition

"We can make the values of $f(x)$ arbitrarily close to $L$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$"

of the limit of $f$ is nothing else than a verbal rephrasing of the standard $varepsilon$-$delta$-definition. Arbitraritly close means that for each $varepsilon > 0$, how big or small it may be, you get

$lvert f(x) - L rvert < varepsilon$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$.

Sufficiently close means that there exists some $delta > 0$ such that

$lvert f(x) - L rvert < varepsilon$ for all $x ne a$ with $lvert x - a rvert < delta$.

Thus "arbitraritly close to $L$" means "for all neighborhoods of $L$" and "sufficiently close to $a$" means "there exists a neighborhood of $a$".

You see that this is not interchangeable.

share|cite|improve this answer

answered Jun 5 at 10:06

Paul FrostPaul Frost

14.7k31035

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if $f(x)$ is fixed from $L$ by specific number isn't this indicate that the limit exist also ?
  $endgroup$
  – Ammar Bamhdi
  Jun 5 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this condition is satisfied, then the limit exists and is equal to $L$.
  $endgroup$
  – Paul Frost
  Jun 5 at 21:50
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3251664%2fwhat-is-the-difference-between-arbitrarily-close-and-sufficiently-close-in-t%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

the limit of $f(x)$, as $x$ approaches $a$, equals $L$” means we can
make the values of $f(x)$ arbitrarily close to $L$

Rephrasing: "as close as we want"

by restricting x to be sufficiently close to $a$ but not equal to $a$.

Rephrasing: "close enough"

---

Or put differently: $L$ is the limit (as $x$ approaches $a$) if we can

• make the distance from $f(x)$ to $L$ as small as we want,


• by only making the distance from $x$ to $a$ small enough.

share|cite|improve this answer

edited Jun 6 at 7:40

answered Jun 5 at 9:27

StackTDStackTD

25.3k2254

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there something wrong with this answer? If the down voter cares to explain, perhaps I can improve it.
  $endgroup$
  – StackTD
  Jun 5 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't downvote, but I suggest replacing "only by" (suggesting "this is the only way") with "by only" (suggesting "using this as our only tool to achieve the desired effect, even though there could be other ways").
  $endgroup$
  – Aasmund Eldhuset
  Jun 5 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AasmundEldhuset Thanks, that's better indeed.
  $endgroup$
  – StackTD
  Jun 6 at 7:40
  

  
  
  
  

add a comment | 

5

$begingroup$

the limit of $f(x)$, as $x$ approaches $a$, equals $L$” means we can
make the values of $f(x)$ arbitrarily close to $L$

Rephrasing: "as close as we want"

by restricting x to be sufficiently close to $a$ but not equal to $a$.

Rephrasing: "close enough"

---

Or put differently: $L$ is the limit (as $x$ approaches $a$) if we can

• make the distance from $f(x)$ to $L$ as small as we want,


• by only making the distance from $x$ to $a$ small enough.

share|cite|improve this answer

edited Jun 6 at 7:40

answered Jun 5 at 9:27

StackTDStackTD

25.3k2254

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there something wrong with this answer? If the down voter cares to explain, perhaps I can improve it.
  $endgroup$
  – StackTD
  Jun 5 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't downvote, but I suggest replacing "only by" (suggesting "this is the only way") with "by only" (suggesting "using this as our only tool to achieve the desired effect, even though there could be other ways").
  $endgroup$
  – Aasmund Eldhuset
  Jun 5 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AasmundEldhuset Thanks, that's better indeed.
  $endgroup$
  – StackTD
  Jun 6 at 7:40
  

  
  
  
  

add a comment | 

$begingroup$

the limit of $f(x)$, as $x$ approaches $a$, equals $L$” means we can
make the values of $f(x)$ arbitrarily close to $L$

Rephrasing: "as close as we want"

by restricting x to be sufficiently close to $a$ but not equal to $a$.

Rephrasing: "close enough"

---

Or put differently: $L$ is the limit (as $x$ approaches $a$) if we can

• make the distance from $f(x)$ to $L$ as small as we want,


• by only making the distance from $x$ to $a$ small enough.

share|cite|improve this answer

edited Jun 6 at 7:40

answered Jun 5 at 9:27

StackTDStackTD

25.3k2254

$endgroup$

share|cite|improve this answer

edited Jun 6 at 7:40

answered Jun 5 at 9:27

StackTDStackTD

25.3k2254

answered Jun 5 at 9:27

StackTDStackTD

25.3k2254

answered Jun 5 at 9:27

StackTDStackTD

25.3k2254

2

$begingroup$

• arbitrarily close: true for any closeness, how large or how small; $forallepsilon$.

$|x-1|>0$ holds for $x$ arbitrarily close to $1$.

• sufficiently close: true under some suitable upper bound; $forallepsilon<u$ where $u$ is dictated by the problem on hand.

$|x-1|<1$ holds for $x$ sufficiently close to $1$.

share|cite|improve this answer

edited Jun 5 at 12:46

answered Jun 5 at 9:40

Yves DaoustYves Daoust

139k880237

$endgroup$

add a comment | 

2

$begingroup$

• arbitrarily close: true for any closeness, how large or how small; $forallepsilon$.

$|x-1|>0$ holds for $x$ arbitrarily close to $1$.

• sufficiently close: true under some suitable upper bound; $forallepsilon<u$ where $u$ is dictated by the problem on hand.

$|x-1|<1$ holds for $x$ sufficiently close to $1$.

share|cite|improve this answer

edited Jun 5 at 12:46

answered Jun 5 at 9:40

Yves DaoustYves Daoust

139k880237

$endgroup$

add a comment | 

$begingroup$

• arbitrarily close: true for any closeness, how large or how small; $forallepsilon$.

$|x-1|>0$ holds for $x$ arbitrarily close to $1$.

• sufficiently close: true under some suitable upper bound; $forallepsilon<u$ where $u$ is dictated by the problem on hand.

$|x-1|<1$ holds for $x$ sufficiently close to $1$.

share|cite|improve this answer

edited Jun 5 at 12:46

answered Jun 5 at 9:40

Yves DaoustYves Daoust

139k880237

$endgroup$

share|cite|improve this answer

edited Jun 5 at 12:46

answered Jun 5 at 9:40

Yves DaoustYves Daoust

139k880237

answered Jun 5 at 9:40

Yves DaoustYves Daoust

139k880237

answered Jun 5 at 9:40

Yves DaoustYves Daoust

139k880237

1

$begingroup$

You do not mention where your function $f$ is defined and which range it has, but let us assume that it is a function $f : mathbb R to mathbb R$. Your definition

"We can make the values of $f(x)$ arbitrarily close to $L$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$"

of the limit of $f$ is nothing else than a verbal rephrasing of the standard $varepsilon$-$delta$-definition. Arbitraritly close means that for each $varepsilon > 0$, how big or small it may be, you get

$lvert f(x) - L rvert < varepsilon$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$.

Sufficiently close means that there exists some $delta > 0$ such that

$lvert f(x) - L rvert < varepsilon$ for all $x ne a$ with $lvert x - a rvert < delta$.

Thus "arbitraritly close to $L$" means "for all neighborhoods of $L$" and "sufficiently close to $a$" means "there exists a neighborhood of $a$".

You see that this is not interchangeable.

share|cite|improve this answer

answered Jun 5 at 10:06

Paul FrostPaul Frost

14.7k31035

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if $f(x)$ is fixed from $L$ by specific number isn't this indicate that the limit exist also ?
  $endgroup$
  – Ammar Bamhdi
  Jun 5 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this condition is satisfied, then the limit exists and is equal to $L$.
  $endgroup$
  – Paul Frost
  Jun 5 at 21:50
  

  
  
  
  

add a comment | 

1

$begingroup$

You do not mention where your function $f$ is defined and which range it has, but let us assume that it is a function $f : mathbb R to mathbb R$. Your definition

"We can make the values of $f(x)$ arbitrarily close to $L$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$"

of the limit of $f$ is nothing else than a verbal rephrasing of the standard $varepsilon$-$delta$-definition. Arbitraritly close means that for each $varepsilon > 0$, how big or small it may be, you get

$lvert f(x) - L rvert < varepsilon$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$.

Sufficiently close means that there exists some $delta > 0$ such that

$lvert f(x) - L rvert < varepsilon$ for all $x ne a$ with $lvert x - a rvert < delta$.

Thus "arbitraritly close to $L$" means "for all neighborhoods of $L$" and "sufficiently close to $a$" means "there exists a neighborhood of $a$".

You see that this is not interchangeable.

share|cite|improve this answer

answered Jun 5 at 10:06

Paul FrostPaul Frost

14.7k31035

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if $f(x)$ is fixed from $L$ by specific number isn't this indicate that the limit exist also ?
  $endgroup$
  – Ammar Bamhdi
  Jun 5 at 13:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this condition is satisfied, then the limit exists and is equal to $L$.
  $endgroup$
  – Paul Frost
  Jun 5 at 21:50
  

  
  
  
  

add a comment | 

$begingroup$

You do not mention where your function $f$ is defined and which range it has, but let us assume that it is a function $f : mathbb R to mathbb R$. Your definition

"We can make the values of $f(x)$ arbitrarily close to $L$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$"

of the limit of $f$ is nothing else than a verbal rephrasing of the standard $varepsilon$-$delta$-definition. Arbitraritly close means that for each $varepsilon > 0$, how big or small it may be, you get

$lvert f(x) - L rvert < varepsilon$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$.

Sufficiently close means that there exists some $delta > 0$ such that

$lvert f(x) - L rvert < varepsilon$ for all $x ne a$ with $lvert x - a rvert < delta$.

Thus "arbitraritly close to $L$" means "for all neighborhoods of $L$" and "sufficiently close to $a$" means "there exists a neighborhood of $a$".

You see that this is not interchangeable.

share|cite|improve this answer

answered Jun 5 at 10:06

Paul FrostPaul Frost

14.7k31035

$endgroup$

share|cite|improve this answer

answered Jun 5 at 10:06

Paul FrostPaul Frost

14.7k31035

answered Jun 5 at 10:06

Paul FrostPaul Frost

14.7k31035

answered Jun 5 at 10:06

Paul FrostPaul Frost

14.7k31035