Indeterminate form and L'Hospital rule












1












$begingroup$



$$
lim_{x to infty}frac{e^x}{x^2}
= lim_{x to infty}frac{e^x}{2x}
= lim_{x to infty}frac{e^x}{2}
= infty
$$




Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $frac{infty}{infty}$ is indeterminate form but at the end, only $infty$ is also indeterminate form.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    $infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 10:58
















1












$begingroup$



$$
lim_{x to infty}frac{e^x}{x^2}
= lim_{x to infty}frac{e^x}{2x}
= lim_{x to infty}frac{e^x}{2}
= infty
$$




Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $frac{infty}{infty}$ is indeterminate form but at the end, only $infty$ is also indeterminate form.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    $infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 10:58














1












1








1





$begingroup$



$$
lim_{x to infty}frac{e^x}{x^2}
= lim_{x to infty}frac{e^x}{2x}
= lim_{x to infty}frac{e^x}{2}
= infty
$$




Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $frac{infty}{infty}$ is indeterminate form but at the end, only $infty$ is also indeterminate form.






calculus







share|cite|improve this question











$endgroup$





$$
lim_{x to infty}frac{e^x}{x^2}
= lim_{x to infty}frac{e^x}{2x}
= lim_{x to infty}frac{e^x}{2}
= infty
$$




Can anybody tell me please why L'hospital's rule been used only two times?
From my point of view, $frac{infty}{infty}$ is indeterminate form but at the end, only $infty$ is also indeterminate form.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 11:06









Namaste

1




1










asked Dec 16 '18 at 10:55









Jobiar HossainJobiar Hossain

414




414












  • $begingroup$
    $infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 10:58


















  • $begingroup$
    $infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
    $endgroup$
    – Viktor Glombik
    Dec 16 '18 at 10:58
















$begingroup$
$infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
$endgroup$
– Viktor Glombik
Dec 16 '18 at 10:58




$begingroup$
$infty$ is not indeterminate form (see here)[en.wikipedia.org/wiki/Indeterminate_form]
$endgroup$
– Viktor Glombik
Dec 16 '18 at 10:58










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



As an alternative note that eventually $e^xge x^3$ and then



$$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



We can prove that also by induction, that is $e^n ge n^3$




  • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

  • induction step: assume true $e^k ge k^3$ for some $k$ then


$$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



and the latter is true indeed



$$2cdot k^3ge (k+1)^3$$



$$k(k^2-3k-3)ge 1, quad kge 4$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Note that $infty$ is not an intermediate form. Simply put :



    $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



    Indeterminate forms are typically considered to be :



    $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I guess you intended to write "indeterminate", not "intermediate"
      $endgroup$
      – jjagmath
      Dec 16 '18 at 11:37










    • $begingroup$
      @jjagmath Thanks!
      $endgroup$
      – Rebellos
      Dec 16 '18 at 11:37











    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%2f3042468%2findeterminate-form-and-lhospital-rule%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









    1












    $begingroup$

    Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



    As an alternative note that eventually $e^xge x^3$ and then



    $$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



    We can prove that also by induction, that is $e^n ge n^3$




    • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

    • induction step: assume true $e^k ge k^3$ for some $k$ then


    $$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



    and the latter is true indeed



    $$2cdot k^3ge (k+1)^3$$



    $$k(k^2-3k-3)ge 1, quad kge 4$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



      As an alternative note that eventually $e^xge x^3$ and then



      $$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



      We can prove that also by induction, that is $e^n ge n^3$




      • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

      • induction step: assume true $e^k ge k^3$ for some $k$ then


      $$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



      and the latter is true indeed



      $$2cdot k^3ge (k+1)^3$$



      $$k(k^2-3k-3)ge 1, quad kge 4$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



        As an alternative note that eventually $e^xge x^3$ and then



        $$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



        We can prove that also by induction, that is $e^n ge n^3$




        • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

        • induction step: assume true $e^k ge k^3$ for some $k$ then


        $$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



        and the latter is true indeed



        $$2cdot k^3ge (k+1)^3$$



        $$k(k^2-3k-3)ge 1, quad kge 4$$






        share|cite|improve this answer











        $endgroup$



        Your application is correct indeed also $frac{e^x}{2x}$ is in the form $frac{infty}{infty}$ while the last one is not an indeterminate form and the limit is indeed infinity.



        As an alternative note that eventually $e^xge x^3$ and then



        $$frac{e^x}{x^2}ge frac{x^3}{x^2}ge xto infty$$



        We can prove that also by induction, that is $e^n ge n^3$




        • base case: $n=1 implies ege 1$, $n=2 implies e^2notge 8$, $n=3 implies e^3notge 27$, $n=4 implies e^4notge 64$,$n=5 implies e^5ge 125$

        • induction step: assume true $e^k ge k^3$ for some $k$ then


        $$e^{k+1}=ecdot e^k ge ecdot k^3ge 2cdot k^3stackrel{?}ge (k+1)^3$$



        and the latter is true indeed



        $$2cdot k^3ge (k+1)^3$$



        $$k(k^2-3k-3)ge 1, quad kge 4$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 11:29

























        answered Dec 16 '18 at 10:57









        gimusigimusi

        93k84494




        93k84494























            1












            $begingroup$

            Note that $infty$ is not an intermediate form. Simply put :



            $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



            Indeterminate forms are typically considered to be :



            $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I guess you intended to write "indeterminate", not "intermediate"
              $endgroup$
              – jjagmath
              Dec 16 '18 at 11:37










            • $begingroup$
              @jjagmath Thanks!
              $endgroup$
              – Rebellos
              Dec 16 '18 at 11:37
















            1












            $begingroup$

            Note that $infty$ is not an intermediate form. Simply put :



            $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



            Indeterminate forms are typically considered to be :



            $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I guess you intended to write "indeterminate", not "intermediate"
              $endgroup$
              – jjagmath
              Dec 16 '18 at 11:37










            • $begingroup$
              @jjagmath Thanks!
              $endgroup$
              – Rebellos
              Dec 16 '18 at 11:37














            1












            1








            1





            $begingroup$

            Note that $infty$ is not an intermediate form. Simply put :



            $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



            Indeterminate forms are typically considered to be :



            $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$






            share|cite|improve this answer











            $endgroup$



            Note that $infty$ is not an intermediate form. Simply put :



            $$lim_{x to infty} frac{e^x}{2} = frac{1}{2}lim_{x to infty} e^x =infty$$



            Indeterminate forms are typically considered to be :



            $$frac{0}{0}, frac{infty}{infty}, 0 cdot infty, 1^infty, infty - infty, 0^infty,infty^0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 16 '18 at 11:37

























            answered Dec 16 '18 at 11:23









            RebellosRebellos

            15.3k31250




            15.3k31250












            • $begingroup$
              I guess you intended to write "indeterminate", not "intermediate"
              $endgroup$
              – jjagmath
              Dec 16 '18 at 11:37










            • $begingroup$
              @jjagmath Thanks!
              $endgroup$
              – Rebellos
              Dec 16 '18 at 11:37


















            • $begingroup$
              I guess you intended to write "indeterminate", not "intermediate"
              $endgroup$
              – jjagmath
              Dec 16 '18 at 11:37










            • $begingroup$
              @jjagmath Thanks!
              $endgroup$
              – Rebellos
              Dec 16 '18 at 11:37
















            $begingroup$
            I guess you intended to write "indeterminate", not "intermediate"
            $endgroup$
            – jjagmath
            Dec 16 '18 at 11:37




            $begingroup$
            I guess you intended to write "indeterminate", not "intermediate"
            $endgroup$
            – jjagmath
            Dec 16 '18 at 11:37












            $begingroup$
            @jjagmath Thanks!
            $endgroup$
            – Rebellos
            Dec 16 '18 at 11:37




            $begingroup$
            @jjagmath Thanks!
            $endgroup$
            – Rebellos
            Dec 16 '18 at 11:37


















            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%2f3042468%2findeterminate-form-and-lhospital-rule%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more