Squeeze theorem laws












2












$begingroup$


I'm trying to understand how the sandwich/ squeeze theorem establishes that the limit of sin theta as theta approaches 0 is 0 and how the limit of cos theta as as theta approaches 0 is 1.



Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.






calculus limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
    $endgroup$
    – trancelocation
    2 hours ago
















2












$begingroup$


I'm trying to understand how the sandwich/ squeeze theorem establishes that the limit of sin theta as theta approaches 0 is 0 and how the limit of cos theta as as theta approaches 0 is 1.



Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.






calculus limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
    $endgroup$
    – trancelocation
    2 hours ago














2












2








2





$begingroup$


I'm trying to understand how the sandwich/ squeeze theorem establishes that the limit of sin theta as theta approaches 0 is 0 and how the limit of cos theta as as theta approaches 0 is 1.



Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.






calculus limits







share|cite|improve this question









$endgroup$




I'm trying to understand how the sandwich/ squeeze theorem establishes that the limit of sin theta as theta approaches 0 is 0 and how the limit of cos theta as as theta approaches 0 is 1.



Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.






calculus limits




calculus limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









JamesJames

163




163








  • 1




    $begingroup$
    Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
    $endgroup$
    – trancelocation
    2 hours ago














  • 1




    $begingroup$
    Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
    $endgroup$
    – trancelocation
    2 hours ago








1




1




$begingroup$
Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
$endgroup$
– trancelocation
2 hours ago




$begingroup$
Are you already equipped with the series representation of $sin theta$ and $cos theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works.
$endgroup$
– trancelocation
2 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.



We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.



For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.



    Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
    But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.



    Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
      $endgroup$
      – Brian Tung
      1 hour ago






    • 1




      $begingroup$
      There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
      $endgroup$
      – Kavi Rama Murthy
      1 hour 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%2f3119710%2fsqueeze-theorem-laws%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 idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.



    We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.



    For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.



      We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.



      For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.



        We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.



        For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.






        share|cite|improve this answer









        $endgroup$



        The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $f$ are lying (sandwiched) between those of $g, h$ and values of $g, h$ tend to a common limit then that of $f$ also tend to the same limit.



        We have the fundamental inequality $|sin x|leq |x|$ for $0<|x|<pi/2$ which is the same as $$-|x|leq sin xleq |x|$$ for $0<|x|<pi/2$ and since both $-|x|$ and $|x|$ tend to $0$ as $xto 0$ it follows that $lim_{xto 0}sin x=0$.



        For $cos x$ the situation is tricky and we need to use a bit of trigonometry. We have $$cos x=1-2sin^2(x/2)$$ and noting the earlier inequality for $sin $ we get $$1-frac{x^2}{2}leq cos xleq 1$$ And by squeeze we get $lim_{xto 0}cos x=1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Paramanand SinghParamanand Singh

        50.2k556163




        50.2k556163























            2












            $begingroup$

            I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.



            Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
            But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.



            Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
              $endgroup$
              – Brian Tung
              1 hour ago






            • 1




              $begingroup$
              There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
              $endgroup$
              – Kavi Rama Murthy
              1 hour ago


















            2












            $begingroup$

            I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.



            Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
            But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.



            Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
              $endgroup$
              – Brian Tung
              1 hour ago






            • 1




              $begingroup$
              There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
              $endgroup$
              – Kavi Rama Murthy
              1 hour ago
















            2












            2








            2





            $begingroup$

            I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.



            Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
            But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.



            Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.






            share|cite|improve this answer











            $endgroup$



            I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers go to a cell, then the prisoner must also end up in the cell. This summarizes the main idea of the theorem.



            Now, we know that when $x$ is sufficiently near to $0$ (i.e. the functions must be defined in a neighbourhood of such a point), $-xle sin xle x$ and this holds for any $x$ sufficiently small. Since limits preserve the linear order, we have that if limit of $sin x$ exists, then $$lim_{xrightarrow0} -xle lim_{xrightarrow0} sin x le lim_{xrightarrow0} x.$$
            But the first and the last limits are $0$, hence limit of $sin x$ exists and is $0$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $sin x$ can't go away.



            Edit: generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 2 hours ago









            LBJFSLBJFS

            1557




            1557








            • 1




              $begingroup$
              Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
              $endgroup$
              – Brian Tung
              1 hour ago






            • 1




              $begingroup$
              There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
              $endgroup$
              – Kavi Rama Murthy
              1 hour ago
















            • 1




              $begingroup$
              Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
              $endgroup$
              – Brian Tung
              1 hour ago






            • 1




              $begingroup$
              There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
              $endgroup$
              – Kavi Rama Murthy
              1 hour ago










            1




            1




            $begingroup$
            Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
            $endgroup$
            – Brian Tung
            1 hour ago




            $begingroup$
            Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-)
            $endgroup$
            – Brian Tung
            1 hour ago




            1




            1




            $begingroup$
            There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
            $endgroup$
            – Kavi Rama Murthy
            1 hour ago






            $begingroup$
            There are some inaccuracies in this answer. The inequalities $-x leq sin , xleq x$ can hold only for $x geq 0$ so one can only take right hand limits.
            $endgroup$
            – Kavi Rama Murthy
            1 hour 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%2f3119710%2fsqueeze-theorem-laws%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

            Liste der Baudenkmale in Friedland (Mecklenburg)

            Image
            In der Liste der Baudenkmale in Friedland sind alle denkmalgeschützten Bauten der Stadt Friedland (Mecklenburg-Vorpommern) und ihrer Ortsteile aufgelistet. Grundlage ist die Veröffentlichung der Denkmalliste des Landkreises Mecklenburgische Seenplatte (Auszug) vom 28. September 2016. Inhaltsverzeichnis 1 Baudenkmale nach Ortsteilen 1.1 Friedland 1.2 Bresewitz 1.3 Brohm 1.4 Cosa 1.5 Dishley 1.6 Dishley-Ausbau 1.7 Eichhorst 1.8 Glienke 1.9 Heinrichshöh 1.10 Heinrichswalde 1.11 Jatzke 1.12 Jatzke-Meierei 1.13 Liepen 1.14 Ramelow 1.15 Schwanbeck 2 Vorgesehener Denkmalbereich 3 Quelle 4 Weblinks Baudenkmale nach Ortsteilen | Friedland | ID Lage Bezeichnung Beschreibung Bild 281 An der Kleinbahn 2 Windmühle mit Umgang 282 Anklamer Straße alter Friedhof mit Portalwand 282 Anklamer Straße drei Erbbegräbnissen und Gräbern u. a. H.A. Riemann 284
            Read more

            Single-Malt-Whisky

            Image
            Fünf verschiedene schottische Single-Malt-Whiskys Als Single-Malt-Whisky oder kürzer Single Malt werden Whiskys bezeichnet, die zwei besondere Bedingungen erfüllen: Sie stammen zum ersten aus einer einzigen Brennerei, sind also kein Verschnitt aus mehreren Whiskysorten (daher Single , deutsch: einzeln). Zum zweiten wird als Getreide ausschließlich gemälzte Gerste verwendet (deswegen Malt , deutsch: Malz). Der Zusatz Single Malt wird als Prädikat verstanden, die so ausgezeichneten Whiskys sind in aller Regel besonders hochwertig. Hinzu kommt ihre besonders lange Lagerung, die Reifungszeit währt oft ein Jahrzehnt und länger. Inhaltsverzeichnis 1 Allgemeines 2 Herkunftsländer 2.1 Schottland 2.2 Irland 2.3 Japan 2.4 Kontinentaleuropa 3 Produktionsschritte 3.1 Mälzen 3.2 Brauen 3.3 Brennen 3.4 Reifung und Finish 3.5 Abfüllung 4 Altersangaben 5 Lagerung von Whiskyflaschen 6 Geschmacksklassifikation 7 Sonstig
            Read more

            Czorneboh

            Image
            Dieser Artikel erläutert den Berg Czorneboh, für die Gottheit siehe Czorneboh (Gottheit). Czorneboh Čornobóh Berggasthof mit Turm Höhe 555,7  m ü. NHN [1] Lage Freistaat Sachsen, Deutschland Gebirge Lausitzer Bergland Dominanz 15,4 km →  Kottmar Koordinaten 51° 7′ 11″  N , 14° 31′ 31″  O 51.119722222222 14.525194444444 555.7 Koordinaten: 51° 7′ 11″  N , 14° 31′ 31″  O Gestein Granodiorit Besonderheiten Aussichtsturm Czorneboh Der Czorneboh , obersorbisch Čornobóh , ist ein Berg zwischen Hochkirch und Cunewalde in der Oberlausitz. Er gehört zu der südöstlich von Bautzen gelegenen Czorneboh-Bergkette und ist mit einer Höhe von 555,7  m ü. NHN [1] der höchste Punkt dieses Ausläufers des Lausitzer Berglandes. Er leitet als Landschaftselement die wellige Granitlandschaft der Lausitzer Vorbergzone ein. Der Gipfel des Czorneboh liegt in der Gemarkung Meschwitz (Gemeinde Hochkirch), die Czornebohbaude jedoch
            Read more