Limit to 0 ambiguity The 2019 Stack Overflow Developer Survey Results Are InNeed help with a...

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

aging parents with no investments

Why did Howard Stark use all the Vibranium they had on a prototype shield?

It's possible to achieve negative score?

What is the motivation for a law requiring 2 parties to consent for recording a conversation

Realistic Alternatives to Dust: What Else Could Feed a Plankton Bloom?

Are USB sockets on wall outlets live all the time, even when the switch is off?

I looked up a future colleague on LinkedIn before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?

In microwave frequencies, do you use a circulator when you need a (near) perfect diode?

Spanish for "widget"

Manuscript was "unsubmitted" because the manuscript was deposited in Arxiv Preprints

"To split hairs" vs "To be pedantic"

Why do UK politicians seemingly ignore opinion polls on Brexit?

Extreme, unacceptable situation and I can't attend work tomorrow morning

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

CiviEvent: Public link for events of a specific type

Is there a name of the flying bionic bird?

Why do I get badly formatted numerical results when I use StringForm?

Why could you hear an Amstrad CPC working?

Monty Hall variation

How can I fix this gap between bookcases I made?

What does Linus Torvalds mean when he says that Git "never ever" tracks a file?

What is the use of option -o in the useradd command?

What are the motivations for publishing new editions of an existing textbook, beyond new discoveries in a field?



Limit to 0 ambiguity



The 2019 Stack Overflow Developer Survey Results Are InNeed help with a limitHow to evaluate indeterminate form of a limitThe limit of $ ln{x} + cot{x}$Limit of a rational function to the power of xEvaluating the limit of $lim_{xtoinfty}(sqrt{frac{x^3}{x+2}}-x)$.Evaluate the following limit without L'HopitalLimit of: $ -x+sqrt{x^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:
$$lim_{x to 0} frac{1}{x}, $$
$$+infty~text{or }-infty$$
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
    3 hours ago






  • 3




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










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    3 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
    3 hours ago






  • 3




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










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    3 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 2 hours ago









user8718165

1167




1167










asked 3 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
    3 hours ago






  • 3




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










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    3 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
    3 hours ago






  • 3




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










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    3 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
3 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
3 hours ago




3




3




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




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












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




$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
3 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, this fact is 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
      2 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 3 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, this fact is 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
              2 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, this fact is 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
              2 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, this fact is 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, this fact is 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 15 mins ago

























            answered 3 hours ago









            Michael RybkinMichael Rybkin

            4,279422




            4,279422












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


















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
















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




            $begingroup$
            Thank you so much!!!
            $endgroup$
            – J.Moh
            2 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

            “%fieldName is a required field.”, in Magento2 REST API Call for GET Method Type The Next...

            Image
            How seriously should I take size and weight limits of hand luggage? Can a virus destroy the BIOS of a modern computer? Can I hook these wires up to find the connection to a dead outlet? How to keep a dark protagonist who wants to keep his humanity dark? How does a dynamic QR code work? My singleton can be called multiple times How to detect integer overflow in C? Can a PhD from a non-TU9 German university become a professor in a TU9 university? How to travel Japan while expressing milk? Is it a bad idea to plug the other end of ESD strap to wall ground? How to show a landlord what we have in s
            Read more

            How to change City field to a dropdown in Checkout step Magento 2Magento 2 : How to change UI field(s)...

            Image
            Does the US political system, in principle, allow for a no-party system? Is it a Cyclops number? "Nobody" knows! Understanding a proof about a set being closed Rationale to prefer local variables over instance variables? Movie: boy escapes the real world and goes to a fantasy world with big furry trolls Why aren't there more Gauls like Obelix? I reported the illegal activity of my boss to his boss. My boss found out. Now I am being punished. What should I do? Solving Linear Matrix Recurrences What is the generally accepted pronunciation of “topoi”? Is it safe to abruplty remove arduino power
            Read more

            夢乃愛華...

            Image
            1995年出生在世人物日本AV女優S1女優埼玉縣出身人物 日本AV女優S1TeStar千葉縣松戶市立旭町中學校松戶市立松戶高等學校淑德大學S1 body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} ゆめの あいか 夢乃 あいか AV女优 别名 安藤琴乃（本名） [1] 民族 大和族 出生 ( 1995-09-18 ) 1995年9月18日 （ 23歲） 出道地点   日本宮城縣 → 千葉縣 → 埼玉縣 活跃年代 2013年 - 血型 B型 体形（2013年 [2] 数值） 身高／体重 149 厘米 / ― 公斤 三围 79 - 52 - 78 cm 罩杯 G 体形备注 巨乳 计量单位换算 身高／体重 4英尺11英寸 / ― 磅 三围 31 - 20 - 31 英寸 活动 拍摄类型 AV 模特儿类型 水着 其他职业 鯛魚燒店員（求學時期） [1] 業餘攝影會模特兒（求學時期） [3] 教学：Template:AV女优/doc 夢乃愛華 （日语： 夢乃 あいか ／ ゆめの あいか   Yumeno Aika ，1995年9月18日 － [1] ）是日本AV女優，现为S1专属女优，所屬經紀公司為TeStar。 目录 1 簡歷 2 演出作品 2.1 成人影片 2.2 形象影片 2.3 綜藝節目 2.4 網上節目 3 其他產品 3.1 雜誌 3.2 寫真集 3.3 網上寫真 4 相關條目 5 參考資料 6 外部連結 簡歷 夢乃愛華出身於宮城縣，有一位姊姊 [3] ，後與家人移居千葉縣，曾就讀當地的松戶市立旭町中學校及松戶市立松戶高等學校 [1] 。她中學一年級時身高已停止增長，而高中初期胸部罩杯亦只有 B，但隨著「初體驗」和均衡飲食帶來的生理影響，三年級畢業那刻罩杯則升至 F [3] 。
            Read more