Uniqueness of limit of sequence and function
$begingroup$
I was doing both proof of uniqueness of limit of a sequence and function and i don't understand why in sequence we take $nu=max{nu',nu''}$ and in function we take $delta=min{delta',delta''}$
I hope it's clear without writing the whole proof of both theorems, thanks.
Let X,Y be metric spaces, $f:Ato Y$ a function and $l',l''in Y$
if $lim_{xto x_0}f(x)=l'$ and $lim_{xto x_0}f(x)=l'$ $implies l'=l''$
Proof.
$forallvarepsilon>0$ $exists delta'>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l')<varepsilon$
$forallvarepsilon>0$ $exists delta''>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l'')<varepsilon$
Let $delta=min{delta',delta''}$. So
$forallvarepsilon>0$ $exists delta>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l')<varepsilon$ $d_Y(f(x),l'')<varepsilon$
Then $forall>0$ $d_Y(l',l'')<2varepsilon$ then $l'=l''$
calculus limits proof-explanation
asked Dec 1 '18 at 19:33
ArchimedessArchimedess
165
$endgroup$
add a comment |
$begingroup$
I was doing both proof of uniqueness of limit of a sequence and function and i don't understand why in sequence we take $nu=max{nu',nu''}$ and in function we take $delta=min{delta',delta''}$
I hope it's clear without writing the whole proof of both theorems, thanks.
Let X,Y be metric spaces, $f:Ato Y$ a function and $l',l''in Y$
if $lim_{xto x_0}f(x)=l'$ and $lim_{xto x_0}f(x)=l'$ $implies l'=l''$
Proof.
$forallvarepsilon>0$ $exists delta'>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l')<varepsilon$
$forallvarepsilon>0$ $exists delta''>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l'')<varepsilon$
Let $delta=min{delta',delta''}$. So
$forallvarepsilon>0$ $exists delta>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l')<varepsilon$ $d_Y(f(x),l'')<varepsilon$
Then $forall>0$ $d_Y(l',l'')<2varepsilon$ then $l'=l''$
calculus limits proof-explanation
asked Dec 1 '18 at 19:33
ArchimedessArchimedess
165
$endgroup$
add a comment |
$begingroup$
I was doing both proof of uniqueness of limit of a sequence and function and i don't understand why in sequence we take $nu=max{nu',nu''}$ and in function we take $delta=min{delta',delta''}$
I hope it's clear without writing the whole proof of both theorems, thanks.
Let X,Y be metric spaces, $f:Ato Y$ a function and $l',l''in Y$
if $lim_{xto x_0}f(x)=l'$ and $lim_{xto x_0}f(x)=l'$ $implies l'=l''$
Proof.
$forallvarepsilon>0$ $exists delta'>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l')<varepsilon$
$forallvarepsilon>0$ $exists delta''>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l'')<varepsilon$
Let $delta=min{delta',delta''}$. So
$forallvarepsilon>0$ $exists delta>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l')<varepsilon$ $d_Y(f(x),l'')<varepsilon$
Then $forall>0$ $d_Y(l',l'')<2varepsilon$ then $l'=l''$
calculus limits proof-explanation
asked Dec 1 '18 at 19:33
ArchimedessArchimedess
165
$endgroup$
I was doing both proof of uniqueness of limit of a sequence and function and i don't understand why in sequence we take $nu=max{nu',nu''}$ and in function we take $delta=min{delta',delta''}$
I hope it's clear without writing the whole proof of both theorems, thanks.
Let X,Y be metric spaces, $f:Ato Y$ a function and $l',l''in Y$
if $lim_{xto x_0}f(x)=l'$ and $lim_{xto x_0}f(x)=l'$ $implies l'=l''$
Proof.
$forallvarepsilon>0$ $exists delta'>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l')<varepsilon$
$forallvarepsilon>0$ $exists delta''>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l'')<varepsilon$
Let $delta=min{delta',delta''}$. So
$forallvarepsilon>0$ $exists delta>0:forall xin A$ $d_X(x,x_0)<delta $ $xneq x_0$ $d_Y(f(x),l')<varepsilon$ $d_Y(f(x),l'')<varepsilon$
Then $forall>0$ $d_Y(l',l'')<2varepsilon$ then $l'=l''$
calculus limits proof-explanation
calculus limits proof-explanation
asked Dec 1 '18 at 19:33
ArchimedessArchimedess
165
asked Dec 1 '18 at 19:33
ArchimedessArchimedess
165
edited Dec 1 '18 at 19:48
Archimedess
asked Dec 1 '18 at 19:33
ArchimedessArchimedess
165
asked Dec 1 '18 at 19:33
ArchimedessArchimedess
165
asked Dec 1 '18 at 19:33
ArchimedessArchimedess
165
165
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, it is clear. When dealing with sequences, we have a condition of the type$$(existsnuinmathbb{N}):ngeqslantnuimplies C_n$$So, if we have a $nu_1$ and a $nu_2$ such that $nimplies C_n$ whenever $ngeqslantnu_1$ and $ngeqslantnu_2$, then we take $nu=max{nu_1,nu_2}$ because$$ngeqslantmax{nu_1,nu_2}iff ngeqslantnu_1text{ and }ngeqslantnu_2.$$But in the case of functions, we have a condition of the type$$lvert x-arvert<deltaimpliescdots$$Before we had $geqslant$ and now we have $<$. So, we consider $min{delta_1,delta_2}$, because$$lvert x-arvert<min{delta_1,delta_2}ifflvert x-arvert<delta_1text{ and }lvert x-arvert<delta_2.$$
edited Dec 1 '18 at 19:42
K Split X
4,19611131
answered Dec 1 '18 at 19:42
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
$begingroup$
Oh thanks, it's pretty obvious now.
$endgroup$
– Archimedess
Dec 1 '18 at 19:53
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021721%2funiqueness-of-limit-of-sequence-and-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, it is clear. When dealing with sequences, we have a condition of the type$$(existsnuinmathbb{N}):ngeqslantnuimplies C_n$$So, if we have a $nu_1$ and a $nu_2$ such that $nimplies C_n$ whenever $ngeqslantnu_1$ and $ngeqslantnu_2$, then we take $nu=max{nu_1,nu_2}$ because$$ngeqslantmax{nu_1,nu_2}iff ngeqslantnu_1text{ and }ngeqslantnu_2.$$But in the case of functions, we have a condition of the type$$lvert x-arvert<deltaimpliescdots$$Before we had $geqslant$ and now we have $<$. So, we consider $min{delta_1,delta_2}$, because$$lvert x-arvert<min{delta_1,delta_2}ifflvert x-arvert<delta_1text{ and }lvert x-arvert<delta_2.$$
edited Dec 1 '18 at 19:42
K Split X
4,19611131
answered Dec 1 '18 at 19:42
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
$begingroup$
Oh thanks, it's pretty obvious now.
$endgroup$
– Archimedess
Dec 1 '18 at 19:53
add a comment |
$begingroup$
Yes, it is clear. When dealing with sequences, we have a condition of the type$$(existsnuinmathbb{N}):ngeqslantnuimplies C_n$$So, if we have a $nu_1$ and a $nu_2$ such that $nimplies C_n$ whenever $ngeqslantnu_1$ and $ngeqslantnu_2$, then we take $nu=max{nu_1,nu_2}$ because$$ngeqslantmax{nu_1,nu_2}iff ngeqslantnu_1text{ and }ngeqslantnu_2.$$But in the case of functions, we have a condition of the type$$lvert x-arvert<deltaimpliescdots$$Before we had $geqslant$ and now we have $<$. So, we consider $min{delta_1,delta_2}$, because$$lvert x-arvert<min{delta_1,delta_2}ifflvert x-arvert<delta_1text{ and }lvert x-arvert<delta_2.$$
edited Dec 1 '18 at 19:42
K Split X
4,19611131
answered Dec 1 '18 at 19:42
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
$begingroup$
Oh thanks, it's pretty obvious now.
$endgroup$
– Archimedess
Dec 1 '18 at 19:53
add a comment |
$begingroup$
Yes, it is clear. When dealing with sequences, we have a condition of the type$$(existsnuinmathbb{N}):ngeqslantnuimplies C_n$$So, if we have a $nu_1$ and a $nu_2$ such that $nimplies C_n$ whenever $ngeqslantnu_1$ and $ngeqslantnu_2$, then we take $nu=max{nu_1,nu_2}$ because$$ngeqslantmax{nu_1,nu_2}iff ngeqslantnu_1text{ and }ngeqslantnu_2.$$But in the case of functions, we have a condition of the type$$lvert x-arvert<deltaimpliescdots$$Before we had $geqslant$ and now we have $<$. So, we consider $min{delta_1,delta_2}$, because$$lvert x-arvert<min{delta_1,delta_2}ifflvert x-arvert<delta_1text{ and }lvert x-arvert<delta_2.$$
edited Dec 1 '18 at 19:42
K Split X
4,19611131
answered Dec 1 '18 at 19:42
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
Yes, it is clear. When dealing with sequences, we have a condition of the type$$(existsnuinmathbb{N}):ngeqslantnuimplies C_n$$So, if we have a $nu_1$ and a $nu_2$ such that $nimplies C_n$ whenever $ngeqslantnu_1$ and $ngeqslantnu_2$, then we take $nu=max{nu_1,nu_2}$ because$$ngeqslantmax{nu_1,nu_2}iff ngeqslantnu_1text{ and }ngeqslantnu_2.$$But in the case of functions, we have a condition of the type$$lvert x-arvert<deltaimpliescdots$$Before we had $geqslant$ and now we have $<$. So, we consider $min{delta_1,delta_2}$, because$$lvert x-arvert<min{delta_1,delta_2}ifflvert x-arvert<delta_1text{ and }lvert x-arvert<delta_2.$$
edited Dec 1 '18 at 19:42
K Split X
4,19611131
answered Dec 1 '18 at 19:42
José Carlos SantosJosé Carlos Santos
154k22123226
edited Dec 1 '18 at 19:42
K Split X
4,19611131
edited Dec 1 '18 at 19:42
K Split X
4,19611131
edited Dec 1 '18 at 19:42
K Split X
4,19611131
4,19611131
answered Dec 1 '18 at 19:42
José Carlos SantosJosé Carlos Santos
154k22123226
answered Dec 1 '18 at 19:42
José Carlos SantosJosé Carlos Santos
154k22123226
answered Dec 1 '18 at 19:42
José Carlos SantosJosé Carlos Santos
154k22123226
154k22123226
$begingroup$
Oh thanks, it's pretty obvious now.
$endgroup$
– Archimedess
Dec 1 '18 at 19:53
add a comment |
$begingroup$
Oh thanks, it's pretty obvious now.
$endgroup$
– Archimedess
Dec 1 '18 at 19:53
$begingroup$
Oh thanks, it's pretty obvious now.
$endgroup$
– Archimedess
Dec 1 '18 at 19:53
$begingroup$
Oh thanks, it's pretty obvious now.
$endgroup$
– Archimedess
Dec 1 '18 at 19:53
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021721%2funiqueness-of-limit-of-sequence-and-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
