Prove using limit definition ($varepsilon - delta$)












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 0:52









mrtaurho

4,05921234




4,05921234










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
















  • 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












1 Answer
1






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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
});


}
});














draft saved

draft discarded


















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









4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















draft saved

draft discarded




















































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.




draft saved


draft discarded














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





















































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







-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more