Prove using limit definition ($varepsilon - delta$)
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be such that $lim_{xto p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $varepsilon-delta$ formalism, that $$lim_{xto p} [f(x)]^2 = L^2$$
My work:
$$lim_{xrightarrow p} [f(x)]^2 = lim_{xto p}f(x) cdotlim_{xto p} f(x) = L cdot L = L^2$$
But that seems too easy, and doesn't involve $varepsilon$ or $delta$. Any other approaches? Please and thank you.
real-analysis limits epsilon-delta
edited Dec 2 '18 at 0:52
mrtaurho
4,05921234
asked Dec 2 '18 at 0:46
Ty JohnsonTy Johnson
323
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be such that $lim_{xto p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $varepsilon-delta$ formalism, that $$lim_{xto p} [f(x)]^2 = L^2$$
My work:
$$lim_{xrightarrow p} [f(x)]^2 = lim_{xto p}f(x) cdotlim_{xto p} f(x) = L cdot L = L^2$$
But that seems too easy, and doesn't involve $varepsilon$ or $delta$. Any other approaches? Please and thank you.
real-analysis limits epsilon-delta
edited Dec 2 '18 at 0:52
mrtaurho
4,05921234
asked Dec 2 '18 at 0:46
Ty JohnsonTy Johnson
323
$endgroup$
2
$begingroup$
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
$endgroup$
– Trevor Gunn
Dec 2 '18 at 0:51
add a comment |
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be such that $lim_{xto p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $varepsilon-delta$ formalism, that $$lim_{xto p} [f(x)]^2 = L^2$$
My work:
$$lim_{xrightarrow p} [f(x)]^2 = lim_{xto p}f(x) cdotlim_{xto p} f(x) = L cdot L = L^2$$
But that seems too easy, and doesn't involve $varepsilon$ or $delta$. Any other approaches? Please and thank you.
real-analysis limits epsilon-delta
edited Dec 2 '18 at 0:52
mrtaurho
4,05921234
asked Dec 2 '18 at 0:46
Ty JohnsonTy Johnson
323
$endgroup$
Let $f:mathbb{R}tomathbb{R}$ be such that $lim_{xto p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $varepsilon-delta$ formalism, that $$lim_{xto p} [f(x)]^2 = L^2$$
My work:
$$lim_{xrightarrow p} [f(x)]^2 = lim_{xto p}f(x) cdotlim_{xto p} f(x) = L cdot L = L^2$$
But that seems too easy, and doesn't involve $varepsilon$ or $delta$. Any other approaches? Please and thank you.
real-analysis limits epsilon-delta
real-analysis limits epsilon-delta
edited Dec 2 '18 at 0:52
mrtaurho
4,05921234
asked Dec 2 '18 at 0:46
Ty JohnsonTy Johnson
323
edited Dec 2 '18 at 0:52
mrtaurho
4,05921234
asked Dec 2 '18 at 0:46
Ty JohnsonTy Johnson
323
edited Dec 2 '18 at 0:52
mrtaurho
4,05921234
edited Dec 2 '18 at 0:52
mrtaurho
4,05921234
edited Dec 2 '18 at 0:52
mrtaurho
4,05921234
4,05921234
asked Dec 2 '18 at 0:46
Ty JohnsonTy Johnson
323
asked Dec 2 '18 at 0:46
Ty JohnsonTy Johnson
323
asked Dec 2 '18 at 0:46
Ty JohnsonTy Johnson
323
323
2
$begingroup$
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
$endgroup$
– Trevor Gunn
Dec 2 '18 at 0:51
add a comment |
2
$begingroup$
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
$endgroup$
– Trevor Gunn
Dec 2 '18 at 0:51
2
2
$begingroup$
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
$endgroup$
– Trevor Gunn
Dec 2 '18 at 0:51
$begingroup$
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
$endgroup$
– Trevor Gunn
Dec 2 '18 at 0:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 '18 at 1:05
FranklinBashFranklinBash
1012
$endgroup$
$begingroup$
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51
$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57
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%2f3022073%2fprove-using-limit-definition-varepsilon-delta%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
$begingroup$
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 '18 at 1:05
FranklinBashFranklinBash
1012
$endgroup$
$begingroup$
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51
$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57
add a comment |
$begingroup$
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 '18 at 1:05
FranklinBashFranklinBash
1012
$endgroup$
$begingroup$
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51
$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57
add a comment |
$begingroup$
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 '18 at 1:05
FranklinBashFranklinBash
1012
$endgroup$
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 '18 at 1:05
FranklinBashFranklinBash
1012
answered Dec 2 '18 at 1:05
FranklinBashFranklinBash
1012
answered Dec 2 '18 at 1:05
FranklinBashFranklinBash
1012
answered Dec 2 '18 at 1:05
FranklinBashFranklinBash
1012
1012
$begingroup$
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51
$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57
add a comment |
$begingroup$
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51
$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57
$begingroup$
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51
$begingroup$
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51
$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57
$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57
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%2f3022073%2fprove-using-limit-definition-varepsilon-delta%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
2
$begingroup$
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
$endgroup$
– Trevor Gunn
Dec 2 '18 at 0:51