Mixing
Find $underset{{x,y}to {0,0}}{text{lim}}exp left(-frac{left| x-yright| }{(x-y)^2}right)$.
$begingroup$
Find $underset{{x,y}to {0,0}}{text{lim}}exp left(-frac{left| x-yright| }{(x-y)^2}right)$ and prove the limit exists by $epsilon - delta$ definition.
real-analysis
asked Jan 18 at 9:45
helios321helios321
56349
$endgroup$
add a comment |
$begingroup$
Find $underset{{x,y}to {0,0}}{text{lim}}exp left(-frac{left| x-yright| }{(x-y)^2}right)$ and prove the limit exists by $epsilon - delta$ definition.
real-analysis
asked Jan 18 at 9:45
helios321helios321
56349
$endgroup$
add a comment |
$begingroup$
Find $underset{{x,y}to {0,0}}{text{lim}}exp left(-frac{left| x-yright| }{(x-y)^2}right)$ and prove the limit exists by $epsilon - delta$ definition.
real-analysis
asked Jan 18 at 9:45
helios321helios321
56349
$endgroup$
Find $underset{{x,y}to {0,0}}{text{lim}}exp left(-frac{left| x-yright| }{(x-y)^2}right)$ and prove the limit exists by $epsilon - delta$ definition.
real-analysis
real-analysis
asked Jan 18 at 9:45
helios321helios321
56349
asked Jan 18 at 9:45
helios321helios321
56349
asked Jan 18 at 9:45
helios321helios321
56349
asked Jan 18 at 9:45
helios321helios321
56349
asked Jan 18 at 9:45
helios321helios321
56349
56349
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The limit is $0$. Let $0<epsilon <1$ and $delta=frac 1 {2ln (frac 1 {epsilon})}$. Note that $frac {|x-y|} {(x-y)^{2}} =frac 1 {|x-y|}$. Can you now show that $e^{-frac {|x-y|} {(x-y)^{2}}}<epsilon$ whenever $|x-y|<delta$? [Use the fact that $|(x,y)|<delta$ implies $|x-y| <|x|+|y| <2delta$].
answered Jan 18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
$endgroup$
$begingroup$
Don't we need to make $sqrt{x^2+y^2}<delta$ for the two dimensional case?
$endgroup$
– helios321
Jan 18 at 9:56
$begingroup$
@helios321 Right, I have made a correction.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:04
$begingroup$
$|x-y| =sqrt{x^2+y^2-2xy}$ ; and $x^2+y^2ge 2|yx| $ (AM-GM); then $|x-y| le sqrt{2(x^2+y^2)} =√2 sqrt{x^2+y^2}$, helps?
$endgroup$
– Peter Szilas
Jan 18 at 10:09
$begingroup$
@PeterSzilas I had typed $|x-y| <delta$ instead of $|(x,y)| <delta$. You don't require any of the inequalities you have stated. It is very elementary.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:14
$begingroup$
KaviRamaMurthy.Fine.Thanks for update.
$endgroup$
– Peter Szilas
Jan 18 at 11:44
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%2f3078031%2ffind-underset-x-y-to-0-0-textlim-exp-left-frac-left-x-y-righ%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
$begingroup$
The limit is $0$. Let $0<epsilon <1$ and $delta=frac 1 {2ln (frac 1 {epsilon})}$. Note that $frac {|x-y|} {(x-y)^{2}} =frac 1 {|x-y|}$. Can you now show that $e^{-frac {|x-y|} {(x-y)^{2}}}<epsilon$ whenever $|x-y|<delta$? [Use the fact that $|(x,y)|<delta$ implies $|x-y| <|x|+|y| <2delta$].
answered Jan 18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
$endgroup$
$begingroup$
Don't we need to make $sqrt{x^2+y^2}<delta$ for the two dimensional case?
$endgroup$
– helios321
Jan 18 at 9:56
$begingroup$
@helios321 Right, I have made a correction.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:04
$begingroup$
$|x-y| =sqrt{x^2+y^2-2xy}$ ; and $x^2+y^2ge 2|yx| $ (AM-GM); then $|x-y| le sqrt{2(x^2+y^2)} =√2 sqrt{x^2+y^2}$, helps?
$endgroup$
– Peter Szilas
Jan 18 at 10:09
$begingroup$
@PeterSzilas I had typed $|x-y| <delta$ instead of $|(x,y)| <delta$. You don't require any of the inequalities you have stated. It is very elementary.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:14
$begingroup$
KaviRamaMurthy.Fine.Thanks for update.
$endgroup$
– Peter Szilas
Jan 18 at 11:44
add a comment |
$begingroup$
The limit is $0$. Let $0<epsilon <1$ and $delta=frac 1 {2ln (frac 1 {epsilon})}$. Note that $frac {|x-y|} {(x-y)^{2}} =frac 1 {|x-y|}$. Can you now show that $e^{-frac {|x-y|} {(x-y)^{2}}}<epsilon$ whenever $|x-y|<delta$? [Use the fact that $|(x,y)|<delta$ implies $|x-y| <|x|+|y| <2delta$].
answered Jan 18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
$endgroup$
$begingroup$
Don't we need to make $sqrt{x^2+y^2}<delta$ for the two dimensional case?
$endgroup$
– helios321
Jan 18 at 9:56
$begingroup$
@helios321 Right, I have made a correction.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:04
$begingroup$
$|x-y| =sqrt{x^2+y^2-2xy}$ ; and $x^2+y^2ge 2|yx| $ (AM-GM); then $|x-y| le sqrt{2(x^2+y^2)} =√2 sqrt{x^2+y^2}$, helps?
$endgroup$
– Peter Szilas
Jan 18 at 10:09
$begingroup$
@PeterSzilas I had typed $|x-y| <delta$ instead of $|(x,y)| <delta$. You don't require any of the inequalities you have stated. It is very elementary.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:14
$begingroup$
KaviRamaMurthy.Fine.Thanks for update.
$endgroup$
– Peter Szilas
Jan 18 at 11:44
add a comment |
$begingroup$
The limit is $0$. Let $0<epsilon <1$ and $delta=frac 1 {2ln (frac 1 {epsilon})}$. Note that $frac {|x-y|} {(x-y)^{2}} =frac 1 {|x-y|}$. Can you now show that $e^{-frac {|x-y|} {(x-y)^{2}}}<epsilon$ whenever $|x-y|<delta$? [Use the fact that $|(x,y)|<delta$ implies $|x-y| <|x|+|y| <2delta$].
answered Jan 18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
$endgroup$
The limit is $0$. Let $0<epsilon <1$ and $delta=frac 1 {2ln (frac 1 {epsilon})}$. Note that $frac {|x-y|} {(x-y)^{2}} =frac 1 {|x-y|}$. Can you now show that $e^{-frac {|x-y|} {(x-y)^{2}}}<epsilon$ whenever $|x-y|<delta$? [Use the fact that $|(x,y)|<delta$ implies $|x-y| <|x|+|y| <2delta$].
answered Jan 18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
edited Jan 18 at 10:12
answered Jan 18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
answered Jan 18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
answered Jan 18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
63.1k42362
63.1k42362
$begingroup$
Don't we need to make $sqrt{x^2+y^2}<delta$ for the two dimensional case?
$endgroup$
– helios321
Jan 18 at 9:56
$begingroup$
@helios321 Right, I have made a correction.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:04
$begingroup$
$|x-y| =sqrt{x^2+y^2-2xy}$ ; and $x^2+y^2ge 2|yx| $ (AM-GM); then $|x-y| le sqrt{2(x^2+y^2)} =√2 sqrt{x^2+y^2}$, helps?
$endgroup$
– Peter Szilas
Jan 18 at 10:09
$begingroup$
@PeterSzilas I had typed $|x-y| <delta$ instead of $|(x,y)| <delta$. You don't require any of the inequalities you have stated. It is very elementary.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:14
$begingroup$
KaviRamaMurthy.Fine.Thanks for update.
$endgroup$
– Peter Szilas
Jan 18 at 11:44
add a comment |
$begingroup$
Don't we need to make $sqrt{x^2+y^2}<delta$ for the two dimensional case?
$endgroup$
– helios321
Jan 18 at 9:56
$begingroup$
@helios321 Right, I have made a correction.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:04
$begingroup$
$|x-y| =sqrt{x^2+y^2-2xy}$ ; and $x^2+y^2ge 2|yx| $ (AM-GM); then $|x-y| le sqrt{2(x^2+y^2)} =√2 sqrt{x^2+y^2}$, helps?
$endgroup$
– Peter Szilas
Jan 18 at 10:09
$begingroup$
@PeterSzilas I had typed $|x-y| <delta$ instead of $|(x,y)| <delta$. You don't require any of the inequalities you have stated. It is very elementary.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:14
$begingroup$
KaviRamaMurthy.Fine.Thanks for update.
$endgroup$
– Peter Szilas
Jan 18 at 11:44
$begingroup$
Don't we need to make $sqrt{x^2+y^2}<delta$ for the two dimensional case?
$endgroup$
– helios321
Jan 18 at 9:56
$begingroup$
Don't we need to make $sqrt{x^2+y^2}<delta$ for the two dimensional case?
$endgroup$
– helios321
Jan 18 at 9:56
$begingroup$
@helios321 Right, I have made a correction.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:04
$begingroup$
@helios321 Right, I have made a correction.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:04
$begingroup$
$|x-y| =sqrt{x^2+y^2-2xy}$ ; and $x^2+y^2ge 2|yx| $ (AM-GM); then $|x-y| le sqrt{2(x^2+y^2)} =√2 sqrt{x^2+y^2}$, helps?
$endgroup$
– Peter Szilas
Jan 18 at 10:09
$begingroup$
$|x-y| =sqrt{x^2+y^2-2xy}$ ; and $x^2+y^2ge 2|yx| $ (AM-GM); then $|x-y| le sqrt{2(x^2+y^2)} =√2 sqrt{x^2+y^2}$, helps?
$endgroup$
– Peter Szilas
Jan 18 at 10:09
$begingroup$
@PeterSzilas I had typed $|x-y| <delta$ instead of $|(x,y)| <delta$. You don't require any of the inequalities you have stated. It is very elementary.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:14
$begingroup$
@PeterSzilas I had typed $|x-y| <delta$ instead of $|(x,y)| <delta$. You don't require any of the inequalities you have stated. It is very elementary.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 10:14
$begingroup$
KaviRamaMurthy.Fine.Thanks for update.
$endgroup$
– Peter Szilas
Jan 18 at 11:44
$begingroup$
KaviRamaMurthy.Fine.Thanks for update.
$endgroup$
– Peter Szilas
Jan 18 at 11:44
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%2f3078031%2ffind-underset-x-y-to-0-0-textlim-exp-left-frac-left-x-y-righ%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
