Limit to 0 ambiguity












1












$begingroup$


I can't determine the limit of such form:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
I tried to get around it, no success.






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
    $endgroup$
    – Dave
    4 hours ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    4 hours ago










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
















1












$begingroup$


I can't determine the limit of such form:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
I tried to get around it, no success.






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
    $endgroup$
    – Dave
    4 hours ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    4 hours ago










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














1












1








1





$begingroup$


I can't determine the limit of such form:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
I tried to get around it, no success.






limits







share|cite|improve this question











$endgroup$




I can't determine the limit of such form:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
I tried to get around it, no success.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









user8718165

1167




1167










asked 4 hours ago









J.MohJ.Moh

505




505








  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
    $endgroup$
    – Dave
    4 hours ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    4 hours ago










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














  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
    $endgroup$
    – Dave
    4 hours ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    4 hours ago










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








1




1




$begingroup$
Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
$endgroup$
– Dave
4 hours ago




$begingroup$
Simply: the limit does not exist. You can say that $$lim_{xto 0^+} frac{1}{x}=infty$$ and $$lim_{xto 0^-} frac{1}{x}=-infty.$$
$endgroup$
– Dave
4 hours ago




3




3




$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
4 hours ago




$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
4 hours ago












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




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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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









$endgroup$





















    3












    $begingroup$

    It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ 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 by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:



    $$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$



    For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (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, $lim_{xto 2}g(x)=-infty$ 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
      3 hours 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









    3












    $begingroup$

    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









    $endgroup$


















      3












      $begingroup$

      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









      $endgroup$
















        3












        3








        3





        $begingroup$

        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 4 hours ago









        MPWMPW

        31.2k12157




        31.2k12157























            3












            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ 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 by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:



            $$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (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, $lim_{xto 2}g(x)=-infty$ 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
              3 hours ago
















            3












            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ 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 by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:



            $$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (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, $lim_{xto 2}g(x)=-infty$ 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
              3 hours ago














            3












            3








            3





            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac{1}{x}$ 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 by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:



            $$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (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, $lim_{xto 2}g(x)=-infty$ 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 $f(x)=frac{1}{x}$ 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 by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:



            $$lim_{xto 0^-}frac{1}{x}=-infty, lim_{xto 0^+}frac{1}{x}=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_{xto 0^-}frac{1}{x}nelim_{xto 0^+}frac{1}{x}$. Thus, $lim_{xto 0}frac{1}{x}=DNE$ (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, $lim_{xto 2}g(x)=-infty$ 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 9 mins ago

























            answered 4 hours ago









            Michael RybkinMichael Rybkin

            4,279422




            4,279422












            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              3 hours ago


















            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              3 hours ago
















            $begingroup$
            Thank you so much!!!
            $endgroup$
            – J.Moh
            3 hours ago




            $begingroup$
            Thank you so much!!!
            $endgroup$
            – J.Moh
            3 hours 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

            Identifying “long and narrow” polygons in with PostGISlength and width of polygonWhy postgis st_overlaps reports Qgis' “avoid intersections” generated polygon as overlapping with others?Adjusting polygons to boundary and filling holesDrawing polygons with fixed area?How to remove spikes in Polygons with PostGISDeleting sliver polygons after difference operation in QGIS?Snapping boundaries in PostGISSplit polygon into parts adding attributes based on underlying polygon in QGISSplitting overlap between polygons and assign to nearest polygon using PostGIS?Expanding polygons and clipping at midpoint?Removing Intersection of Buffers in Same Layers

            Image
            Do I have to worry about players making “bad” choices on level up? Is thermodynamics only applicable to systems in equilibrium? gnu parallel how to use with ffmpeg Has any spacecraft ever had the ability to directly communicate with civilian air traffic control? Reverse the word in a string with the same order in javascript Is creating your own "experiment" considered cheating during a physics exam? How to stop co-workers from teasing me because I know Russian? What is a Recurrent Neural Network? Any examples of headwear for races with animal ears? Confusion about capacitors TikZ how to make supply and demand arrows for nodes? Does a creature that is immune to a condition still make a saving throw? How to verbalise code in Mathematica? Upright [...] in italics quotation Cannot populate data in lightning data table Stark VS Thanos Binary Numbers Magic Trick In gnome-terminal only 2 out of 3 zoom keys work You look catfish vs You look like a
            Read more

            Masuk log Menu navigasi

            Image
            JawaSundaBanyumasanBugisAcehBanjarMinangMelayuDeutschEnglishEspañolFrançaisItalianoNederlands (function()var node=document.getElementById("mw-dismissablenotice-anonplace");if(node)node.outerHTML="u003Cdiv class="mw-dismissable-notice"u003Eu003Cdiv class="mw-dismissable-notice-close"u003E[u003Ca tabindex="0" role="button"u003Etutupu003C/au003E]u003C/divu003Eu003Cdiv class="mw-dismissable-notice-body"u003Eu003Cdiv id="localNotice" lang="id" dir="ltr"u003Eu003C/divu003Eu003C/divu003Eu003C/divu003E";()); Masuk log Loncat ke navigasi Loncat ke pencarian Bahasa: Jawa | Sunda | Banyumasan | Bugis | Aceh | Banjar | Minang | Melayu | Deutsch | English | Español | Français | Italiano | Nederlands Nama pengguna Kata sandi   Biarkan saya tetap masuk Masuk log Bantuan masuk log Lupa kata sandi? Belum punya akun?Gabung Wikipedia Diperoleh dari "https:/
            Read more

            อาณาจักร (ชีววิทยา) ดูเพิ่ม อ้างอิง รายการเลือกการนำทาง10.1086/39456810.5962/bhl.title.447410.1126/science.163.3863.150576276010.1007/BF01796092408502"Phylogenetic structure of the prokaryotic domain: the primary kingdoms"10.1073/pnas.74.11.5088432104270744"Towards a natural system of organisms: proposal for the domains Archaea, Bacteria, and Eucarya"1990PNAS...87.4576W10.1073/pnas.87.12.4576541592112744PubMedJump the queueexpand by handPubMedJump the queueexpand by handPubMedJump the queueexpand by hand"A revised six-kingdom system of life"10.1111/j.1469-185X.1998.tb00030.x9809012"Only six kingdoms of life"10.1098/rspb.2004.2705169172415306349"Kingdoms Protozoa and Chromista and the eozoan root of the eukaryotic tree"10.1098/rsbl.2009.0948288006020031978กเพิ่มข้อมูล

            Image
            Pages using PMID magic linksอาณาจักร (ชีววิทยา)บทความเกี่ยวกับ ชีววิทยา ที่ยังไม่สมบูรณ์ ชีววิทยาอนุกรมวิธานอาณาจักรโดเมน (function()var node=document.getElementById("mw-dismissablenotice-anonplace");if(node)node.outerHTML="u003Cdiv class="mw-dismissable-notice"u003Eu003Cdiv class="mw-dismissable-notice-close"u003E[u003Ca tabindex="0" role="button"u003Eปิดu003C/au003E]u003C/divu003Eu003Cdiv class="mw-dismissable-notice-body"u003Eu003Cdiv id="localNotice" lang="th" dir="ltr"u003Eu003Cp class="mw-empty-elt"u003Eu003C/pu003Eu003C/divu003Eu003C/divu003Eu003C/divu003E";()); อาณาจักร (ชีววิทยา) จากวิกิพีเดีย สารานุกรมเสรี ไปยังการนำทาง ไปยังการค้นหา แปดลำดับขั้นสำคัญ ๆ ของการจำแนกชั้นทางวิทยาศาสตร์ ในหนึ่งโดเมนประกอบด้วยหนึ่งอาณาจักรขึ้นไป ตามแผนภาพไม่ได้แสดงลำดับขั้นที่อยู่อื่น ๆ ที่อยู่ระหว่างนี้ ในชีววิทยา, อาณาจักร [Kingdom] เป็นระดับหรือหม
            Read more