2 path test for non-existenceTwo variable limits via paths - are there pathalogical examples?Showing that a limit exists in $mathbbR^n$Showing limit does not exist using two-path testPath test versus iterated limits for proving existence of a limitA rigorous statement and proof of the two-path testChoosing path to show limit does not existTwo Path Test for finding limit.Limit exist for two path testHow to show a multivariable limit at a point $textbfx$ of $mathbbR^N$ doesn't exist by the path method?Two-path test proof
Why Shazam when there is already Superman?
Doesn't the system of the Supreme Court oppose justice?
Can I say "fingers" when referring to toes?
Pre-mixing cryogenic fuels and using only one fuel tank
Mimic lecturing on blackboard, facing audience
What to do when eye contact makes your coworker uncomfortable?
Quoting Keynes in a lecture
Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?
How could a planet have erratic days?
Does Doodling or Improvising on the Piano Have Any Benefits?
How do you make your own symbol when Detexify fails?
Will number of steps recorded on FitBit/any fitness tracker add up distance in PokemonGo?
How to create a paid keyvalue store
Make a Bowl of Alphabet Soup
Why is it that I can sometimes guess the next note?
What's the name of the logical fallacy where a debater extends a statement far beyond the original statement to make it true?
Do we have to expect a queue for the shuttle from Watford Junction to Harry Potter Studio?
Which was the first story featuring espers?
When were female captains banned from Starfleet?
Biological Blimps: Propulsion
Does the reader need to like the PoV character?
Were Persian-Median kings illiterate?
What is the English pronunciation of "pain au chocolat"?
Is it allowed to activate the ability of multiple planeswalkers in a single turn?
2 path test for non-existence
Two variable limits via paths - are there pathalogical examples?Showing that a limit exists in $mathbbR^n$Showing limit does not exist using two-path testPath test versus iterated limits for proving existence of a limitA rigorous statement and proof of the two-path testChoosing path to show limit does not existTwo Path Test for finding limit.Limit exist for two path testHow to show a multivariable limit at a point $textbfx$ of $mathbbR^N$ doesn't exist by the path method?Two-path test proof
$begingroup$
Find the limit or show that it does not exist:
$lim _left(x,:yright)to left(1,:-1right)left(fracxy+1x^2-y^2right)$
For this question i have used 2 different paths:
Path 1: $x=0$,
where,
$lim _left(yright)to left(-1right)left(frac1-y^2right) = -1$
Path 2: $y=0$,
where,
$lim _left(xright)to left(1right)left(frac1x^2right) = 1$
This yielded 2 different limits, hence the limit does not exist.
Is this the right approach, and if so, can I use any path to my liking?
calculus limits multivariable-calculus
edited Mar 14 at 17:55
Yanko
7,9651830
asked Mar 14 at 17:40
Ashwin selvakumarAshwin selvakumar
84
$endgroup$
add a comment |
$begingroup$
Find the limit or show that it does not exist:
$lim _left(x,:yright)to left(1,:-1right)left(fracxy+1x^2-y^2right)$
For this question i have used 2 different paths:
Path 1: $x=0$,
where,
$lim _left(yright)to left(-1right)left(frac1-y^2right) = -1$
Path 2: $y=0$,
where,
$lim _left(xright)to left(1right)left(frac1x^2right) = 1$
This yielded 2 different limits, hence the limit does not exist.
Is this the right approach, and if so, can I use any path to my liking?
calculus limits multivariable-calculus
edited Mar 14 at 17:55
Yanko
7,9651830
asked Mar 14 at 17:40
Ashwin selvakumarAshwin selvakumar
84
$endgroup$
$begingroup$
There's a problem with your path. You must get $x$ and $y$ to $1$ and $-1$. You can't put $x=0$, you can try $x=1$ instead.
$endgroup$
– Yanko
Mar 14 at 17:54
$begingroup$
@Yanko , why can't i use x = 0?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:58
$begingroup$
@Yanko so i have to use paths x=1 and y= -1?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:59
$begingroup$
You don't have to. But you can. You must choose paths that tends to $(1,-1)$. The path $(0,y)$ as $yrightarrow -1$ goes to $(0,-1)$ so you can't use it.
$endgroup$
– Yanko
Mar 14 at 19:09
add a comment |
$begingroup$
Find the limit or show that it does not exist:
$lim _left(x,:yright)to left(1,:-1right)left(fracxy+1x^2-y^2right)$
For this question i have used 2 different paths:
Path 1: $x=0$,
where,
$lim _left(yright)to left(-1right)left(frac1-y^2right) = -1$
Path 2: $y=0$,
where,
$lim _left(xright)to left(1right)left(frac1x^2right) = 1$
This yielded 2 different limits, hence the limit does not exist.
Is this the right approach, and if so, can I use any path to my liking?
calculus limits multivariable-calculus
edited Mar 14 at 17:55
Yanko
7,9651830
asked Mar 14 at 17:40
Ashwin selvakumarAshwin selvakumar
84
$endgroup$
Find the limit or show that it does not exist:
$lim _left(x,:yright)to left(1,:-1right)left(fracxy+1x^2-y^2right)$
For this question i have used 2 different paths:
Path 1: $x=0$,
where,
$lim _left(yright)to left(-1right)left(frac1-y^2right) = -1$
Path 2: $y=0$,
where,
$lim _left(xright)to left(1right)left(frac1x^2right) = 1$
This yielded 2 different limits, hence the limit does not exist.
Is this the right approach, and if so, can I use any path to my liking?
calculus limits multivariable-calculus
calculus limits multivariable-calculus
edited Mar 14 at 17:55
Yanko
7,9651830
asked Mar 14 at 17:40
Ashwin selvakumarAshwin selvakumar
84
edited Mar 14 at 17:55
Yanko
7,9651830
asked Mar 14 at 17:40
Ashwin selvakumarAshwin selvakumar
84
edited Mar 14 at 17:55
Yanko
7,9651830
edited Mar 14 at 17:55
Yanko
7,9651830
edited Mar 14 at 17:55
Yanko
7,9651830
7,9651830
asked Mar 14 at 17:40
Ashwin selvakumarAshwin selvakumar
84
asked Mar 14 at 17:40
Ashwin selvakumarAshwin selvakumar
84
asked Mar 14 at 17:40
Ashwin selvakumarAshwin selvakumar
84
84
$begingroup$
There's a problem with your path. You must get $x$ and $y$ to $1$ and $-1$. You can't put $x=0$, you can try $x=1$ instead.
$endgroup$
– Yanko
Mar 14 at 17:54
$begingroup$
@Yanko , why can't i use x = 0?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:58
$begingroup$
@Yanko so i have to use paths x=1 and y= -1?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:59
$begingroup$
You don't have to. But you can. You must choose paths that tends to $(1,-1)$. The path $(0,y)$ as $yrightarrow -1$ goes to $(0,-1)$ so you can't use it.
$endgroup$
– Yanko
Mar 14 at 19:09
add a comment |
$begingroup$
There's a problem with your path. You must get $x$ and $y$ to $1$ and $-1$. You can't put $x=0$, you can try $x=1$ instead.
$endgroup$
– Yanko
Mar 14 at 17:54
$begingroup$
@Yanko , why can't i use x = 0?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:58
$begingroup$
@Yanko so i have to use paths x=1 and y= -1?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:59
$begingroup$
You don't have to. But you can. You must choose paths that tends to $(1,-1)$. The path $(0,y)$ as $yrightarrow -1$ goes to $(0,-1)$ so you can't use it.
$endgroup$
– Yanko
Mar 14 at 19:09
$begingroup$
There's a problem with your path. You must get $x$ and $y$ to $1$ and $-1$. You can't put $x=0$, you can try $x=1$ instead.
$endgroup$
– Yanko
Mar 14 at 17:54
$begingroup$
There's a problem with your path. You must get $x$ and $y$ to $1$ and $-1$. You can't put $x=0$, you can try $x=1$ instead.
$endgroup$
– Yanko
Mar 14 at 17:54
$begingroup$
@Yanko , why can't i use x = 0?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:58
$begingroup$
@Yanko , why can't i use x = 0?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:58
$begingroup$
@Yanko so i have to use paths x=1 and y= -1?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:59
$begingroup$
@Yanko so i have to use paths x=1 and y= -1?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:59
$begingroup$
You don't have to. But you can. You must choose paths that tends to $(1,-1)$. The path $(0,y)$ as $yrightarrow -1$ goes to $(0,-1)$ so you can't use it.
$endgroup$
– Yanko
Mar 14 at 19:09
$begingroup$
You don't have to. But you can. You must choose paths that tends to $(1,-1)$. The path $(0,y)$ as $yrightarrow -1$ goes to $(0,-1)$ so you can't use it.
$endgroup$
– Yanko
Mar 14 at 19:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're right. The sequence doesn't converge, however your reasoning is wrong.
The fact that $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists tells you nothing about how $f$ behaves around $(0,-1)$ and therefore you putting $x=0$ makes no sense.
If $(x,y)rightarrow (1,-1)$ then so is $(1,y)$. Therefore if (by contradiction) $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists so does $lim_yrightarrow -1 f(1,y)$ and both limits must be equal.
$f(1,y) =fracy+11-y^2 = frac1+y(1-y)(1+y) = frac11-y$ whenever $ynot = -1$. Therefore $lim_yrightarrow -1 f(1,y) = frac12$.
Similar calculation would yield that $lim_xrightarrow 1 f(x,-1) = -frac12$.
However the existence of $lim_(x,y)rightarrow(1,-1) f(x,y)$ implies that
$$-frac12 =lim_xrightarrow 1 f(x,-1) = lim_(x,y)rightarrow(1,-1) f(x,y) = lim_yrightarrow -1 f(1,y)= frac12$$
which is a contradiction.
answered Mar 14 at 19:56
YankoYanko
7,9651830
$endgroup$
$begingroup$
Thank you, your explanation makes sense:) @Yanko
$endgroup$
– Ashwin selvakumar
Mar 15 at 18:34
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%2f3148295%2f2-path-test-for-non-existence%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
$begingroup$
You're right. The sequence doesn't converge, however your reasoning is wrong.
The fact that $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists tells you nothing about how $f$ behaves around $(0,-1)$ and therefore you putting $x=0$ makes no sense.
If $(x,y)rightarrow (1,-1)$ then so is $(1,y)$. Therefore if (by contradiction) $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists so does $lim_yrightarrow -1 f(1,y)$ and both limits must be equal.
$f(1,y) =fracy+11-y^2 = frac1+y(1-y)(1+y) = frac11-y$ whenever $ynot = -1$. Therefore $lim_yrightarrow -1 f(1,y) = frac12$.
Similar calculation would yield that $lim_xrightarrow 1 f(x,-1) = -frac12$.
However the existence of $lim_(x,y)rightarrow(1,-1) f(x,y)$ implies that
$$-frac12 =lim_xrightarrow 1 f(x,-1) = lim_(x,y)rightarrow(1,-1) f(x,y) = lim_yrightarrow -1 f(1,y)= frac12$$
which is a contradiction.
answered Mar 14 at 19:56
YankoYanko
7,9651830
$endgroup$
$begingroup$
Thank you, your explanation makes sense:) @Yanko
$endgroup$
– Ashwin selvakumar
Mar 15 at 18:34
add a comment |
$begingroup$
You're right. The sequence doesn't converge, however your reasoning is wrong.
The fact that $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists tells you nothing about how $f$ behaves around $(0,-1)$ and therefore you putting $x=0$ makes no sense.
If $(x,y)rightarrow (1,-1)$ then so is $(1,y)$. Therefore if (by contradiction) $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists so does $lim_yrightarrow -1 f(1,y)$ and both limits must be equal.
$f(1,y) =fracy+11-y^2 = frac1+y(1-y)(1+y) = frac11-y$ whenever $ynot = -1$. Therefore $lim_yrightarrow -1 f(1,y) = frac12$.
Similar calculation would yield that $lim_xrightarrow 1 f(x,-1) = -frac12$.
However the existence of $lim_(x,y)rightarrow(1,-1) f(x,y)$ implies that
$$-frac12 =lim_xrightarrow 1 f(x,-1) = lim_(x,y)rightarrow(1,-1) f(x,y) = lim_yrightarrow -1 f(1,y)= frac12$$
which is a contradiction.
answered Mar 14 at 19:56
YankoYanko
7,9651830
$endgroup$
$begingroup$
Thank you, your explanation makes sense:) @Yanko
$endgroup$
– Ashwin selvakumar
Mar 15 at 18:34
add a comment |
$begingroup$
You're right. The sequence doesn't converge, however your reasoning is wrong.
The fact that $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists tells you nothing about how $f$ behaves around $(0,-1)$ and therefore you putting $x=0$ makes no sense.
If $(x,y)rightarrow (1,-1)$ then so is $(1,y)$. Therefore if (by contradiction) $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists so does $lim_yrightarrow -1 f(1,y)$ and both limits must be equal.
$f(1,y) =fracy+11-y^2 = frac1+y(1-y)(1+y) = frac11-y$ whenever $ynot = -1$. Therefore $lim_yrightarrow -1 f(1,y) = frac12$.
Similar calculation would yield that $lim_xrightarrow 1 f(x,-1) = -frac12$.
However the existence of $lim_(x,y)rightarrow(1,-1) f(x,y)$ implies that
$$-frac12 =lim_xrightarrow 1 f(x,-1) = lim_(x,y)rightarrow(1,-1) f(x,y) = lim_yrightarrow -1 f(1,y)= frac12$$
which is a contradiction.
answered Mar 14 at 19:56
YankoYanko
7,9651830
$endgroup$
You're right. The sequence doesn't converge, however your reasoning is wrong.
The fact that $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists tells you nothing about how $f$ behaves around $(0,-1)$ and therefore you putting $x=0$ makes no sense.
If $(x,y)rightarrow (1,-1)$ then so is $(1,y)$. Therefore if (by contradiction) $lim_(x,y)rightarrow(1,-1) f(x,y)$ exists so does $lim_yrightarrow -1 f(1,y)$ and both limits must be equal.
$f(1,y) =fracy+11-y^2 = frac1+y(1-y)(1+y) = frac11-y$ whenever $ynot = -1$. Therefore $lim_yrightarrow -1 f(1,y) = frac12$.
Similar calculation would yield that $lim_xrightarrow 1 f(x,-1) = -frac12$.
However the existence of $lim_(x,y)rightarrow(1,-1) f(x,y)$ implies that
$$-frac12 =lim_xrightarrow 1 f(x,-1) = lim_(x,y)rightarrow(1,-1) f(x,y) = lim_yrightarrow -1 f(1,y)= frac12$$
which is a contradiction.
answered Mar 14 at 19:56
YankoYanko
7,9651830
answered Mar 14 at 19:56
YankoYanko
7,9651830
answered Mar 14 at 19:56
YankoYanko
7,9651830
answered Mar 14 at 19:56
YankoYanko
7,9651830
7,9651830
$begingroup$
Thank you, your explanation makes sense:) @Yanko
$endgroup$
– Ashwin selvakumar
Mar 15 at 18:34
add a comment |
$begingroup$
Thank you, your explanation makes sense:) @Yanko
$endgroup$
– Ashwin selvakumar
Mar 15 at 18:34
$begingroup$
Thank you, your explanation makes sense:) @Yanko
$endgroup$
– Ashwin selvakumar
Mar 15 at 18:34
$begingroup$
Thank you, your explanation makes sense:) @Yanko
$endgroup$
– Ashwin selvakumar
Mar 15 at 18:34
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%2f3148295%2f2-path-test-for-non-existence%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
$begingroup$
There's a problem with your path. You must get $x$ and $y$ to $1$ and $-1$. You can't put $x=0$, you can try $x=1$ instead.
$endgroup$
– Yanko
Mar 14 at 17:54
$begingroup$
@Yanko , why can't i use x = 0?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:58
$begingroup$
@Yanko so i have to use paths x=1 and y= -1?
$endgroup$
– Ashwin selvakumar
Mar 14 at 17:59
$begingroup$
You don't have to. But you can. You must choose paths that tends to $(1,-1)$. The path $(0,y)$ as $yrightarrow -1$ goes to $(0,-1)$ so you can't use it.
$endgroup$
– Yanko
Mar 14 at 19:09