Limit to 0 ambiguity The 2019 Stack Overflow Developer Survey Results Are InLimit Calculation - Sequences at infinityNeed help with a limitStrategy for tackling $lim_ntoinftyfracn(ln n)^-p.$Evaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Finding a basic limitCan the limit of a polynomial involving infinity be finite?Limit of: $ -x+sqrtx^2+x $ for $ xtoinfty $Limit with integral and powerWhat is the result of the following limit?Determining if a multivariable limit exists

Inversion Puzzle

If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?

How was Skylab's orbit inclination chosen?

What do hard-Brexiteers want with respect to the Irish border?

Why Did Howard Stark Use All The Vibranium They Had On A Prototype Shield?

Inline version of a function returns different value than non-inline version

Is it possible for the two major parties in the UK to form a coalition with each other instead of a much smaller party?

Spanish for "widget"

Does a dangling wire really electrocute me if I'm standing in water?

How to change the limits of integration

"Riffle" two strings

How to deal with fear of taking dependencies

Is there a name of the flying bionic bird?

Deadlock Graph and Interpretation, solution to avoid

Are there any other methods to apply to solving simultaneous equations?

How are circuits which use complex ICs normally simulated?

Pristine Bit Checking

Unbreakable Formation vs. Cry of the Carnarium

On the insanity of kings as an argument against monarchy

Could JWST stay at L2 "forever"?

Why is the maximum length of OpenWrt’s root password 8 characters?

A poker game description that does not feel gimmicky

Why is my p-value correlated to difference between means in two sample tests?

How to make payment on the internet without leaving a money trail?



Limit to 0 ambiguity



The 2019 Stack Overflow Developer Survey Results Are InLimit Calculation - Sequences at infinityNeed help with a limitStrategy for tackling $lim_ntoinftyfracn(ln n)^-p.$Evaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Finding a basic limitCan the limit of a polynomial involving infinity be finite?Limit of: $ -x+sqrtx^2+x $ for $ xtoinfty $Limit with integral and powerWhat is the result of the following limit?Determining if a multivariable limit exists










1












$begingroup$


I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
    $endgroup$
    – Dave
    1 hour ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    1 hour ago










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    1 hour ago















1












$begingroup$


I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
    $endgroup$
    – Dave
    1 hour ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    1 hour ago










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    1 hour ago













1












1








1





$begingroup$


I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.










share|cite|improve this question











$endgroup$




I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.







limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 mins ago









user8718165

1167




1167










asked 1 hour ago









J.MohJ.Moh

475




475







  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
    $endgroup$
    – Dave
    1 hour ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    1 hour ago










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    1 hour ago












  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
    $endgroup$
    – Dave
    1 hour ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    1 hour ago










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    1 hour ago







1




1




$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
1 hour ago




$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
1 hour ago




3




3




$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago




$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago












$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour ago




$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



    enter image description here



    As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



    $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



    For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



    Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thank you so much!!!
      $endgroup$
      – J.Moh
      36 mins ago











    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%2f3181778%2flimit-to-0-ambiguity%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.






        share|cite|improve this answer









        $endgroup$



        The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        MPWMPW

        31.2k12157




        31.2k12157





















            2












            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



            enter image description here



            As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



            $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



            Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              36 mins ago















            2












            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



            enter image description here



            As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



            $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



            Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              36 mins ago













            2












            2








            2





            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



            enter image description here



            As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



            $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



            Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.






            share|cite|improve this answer











            $endgroup$



            It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



            enter image description here



            As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



            $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



            Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 47 mins ago









            J. W. Tanner

            4,6441320




            4,6441320










            answered 1 hour ago









            Michael RybkinMichael Rybkin

            4,259422




            4,259422











            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              36 mins ago
















            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              36 mins ago















            $begingroup$
            Thank you so much!!!
            $endgroup$
            – J.Moh
            36 mins ago




            $begingroup$
            Thank you so much!!!
            $endgroup$
            – J.Moh
            36 mins ago

















            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%2f3181778%2flimit-to-0-ambiguity%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

            Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

            2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

            Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e