Limit of rational function solving












1












$begingroup$


I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?



$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$






algebra-precalculus limits rational-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    So $1=1*infty$? I never knew that!
    $endgroup$
    – Lord Shark the Unknown
    May 25 '17 at 10:49










  • $begingroup$
    I cannot follow your every single step!!
    $endgroup$
    – velut luna
    May 25 '17 at 10:52










  • $begingroup$
    Please check edited answer. Is that correct?
    $endgroup$
    – TeodorKolev
    May 25 '17 at 10:56






  • 1




    $begingroup$
    The editing answer is exactly correct
    $endgroup$
    – Noor Aslam
    May 25 '17 at 11:02






  • 2




    $begingroup$
    The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
    $endgroup$
    – B. Mehta
    May 25 '17 at 11:23
















1












$begingroup$


I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?



$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$






algebra-precalculus limits rational-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    So $1=1*infty$? I never knew that!
    $endgroup$
    – Lord Shark the Unknown
    May 25 '17 at 10:49










  • $begingroup$
    I cannot follow your every single step!!
    $endgroup$
    – velut luna
    May 25 '17 at 10:52










  • $begingroup$
    Please check edited answer. Is that correct?
    $endgroup$
    – TeodorKolev
    May 25 '17 at 10:56






  • 1




    $begingroup$
    The editing answer is exactly correct
    $endgroup$
    – Noor Aslam
    May 25 '17 at 11:02






  • 2




    $begingroup$
    The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
    $endgroup$
    – B. Mehta
    May 25 '17 at 11:23














1












1








1





$begingroup$


I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?



$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$






algebra-precalculus limits rational-functions







share|cite|improve this question











$endgroup$




I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?



$$lim_{xto infty}frac{x^3-2x-1}{x^5-2x-1}$$
Step 1:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)}$$
Step 2:
$$lim_{xtoinfty}frac{x^5left(frac{1}{x^2}-frac{2}{x^4}-frac{1}{x^5}right)}{x^5left(1-frac{2}{x^4}-frac{1}{x^5}right)} = frac{0-0-0}{1-0-0}=frac{0}{1}=0$$






algebra-precalculus limits rational-functions




algebra-precalculus limits rational-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 25 '17 at 11:30







TeodorKolev

















asked May 25 '17 at 10:47









TeodorKolevTeodorKolev

207212




207212












  • $begingroup$
    So $1=1*infty$? I never knew that!
    $endgroup$
    – Lord Shark the Unknown
    May 25 '17 at 10:49










  • $begingroup$
    I cannot follow your every single step!!
    $endgroup$
    – velut luna
    May 25 '17 at 10:52










  • $begingroup$
    Please check edited answer. Is that correct?
    $endgroup$
    – TeodorKolev
    May 25 '17 at 10:56






  • 1




    $begingroup$
    The editing answer is exactly correct
    $endgroup$
    – Noor Aslam
    May 25 '17 at 11:02






  • 2




    $begingroup$
    The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
    $endgroup$
    – B. Mehta
    May 25 '17 at 11:23


















  • $begingroup$
    So $1=1*infty$? I never knew that!
    $endgroup$
    – Lord Shark the Unknown
    May 25 '17 at 10:49










  • $begingroup$
    I cannot follow your every single step!!
    $endgroup$
    – velut luna
    May 25 '17 at 10:52










  • $begingroup$
    Please check edited answer. Is that correct?
    $endgroup$
    – TeodorKolev
    May 25 '17 at 10:56






  • 1




    $begingroup$
    The editing answer is exactly correct
    $endgroup$
    – Noor Aslam
    May 25 '17 at 11:02






  • 2




    $begingroup$
    The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
    $endgroup$
    – B. Mehta
    May 25 '17 at 11:23
















$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49




$begingroup$
So $1=1*infty$? I never knew that!
$endgroup$
– Lord Shark the Unknown
May 25 '17 at 10:49












$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52




$begingroup$
I cannot follow your every single step!!
$endgroup$
– velut luna
May 25 '17 at 10:52












$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56




$begingroup$
Please check edited answer. Is that correct?
$endgroup$
– TeodorKolev
May 25 '17 at 10:56




1




1




$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02




$begingroup$
The editing answer is exactly correct
$endgroup$
– Noor Aslam
May 25 '17 at 11:02




2




2




$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23




$begingroup$
The answer as it stands now certainly isn't correct, the numerator in step 1 has two incorrect terms, namely $-2/x^2$ and $-1/x^3$
$endgroup$
– B. Mehta
May 25 '17 at 11:23










1 Answer
1






active

oldest

votes


















1












$begingroup$

The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
{text{if};;limlimits_{x to a} gleft( x right) ne 0}$$ to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



To Sum It Up



Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



$bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



$bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



$bullet$ If $n<m$ then the limit is $0$.



$bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



Note that this method is only for limits to infinity






share|cite|improve this answer









$endgroup$













    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%2f2296069%2flimit-of-rational-function-solving%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
    The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



    For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



    Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
    {text{if};;limlimits_{x to a} gleft( x right) ne 0}$$     to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



    To Sum It Up



    Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

    Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



    $bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



    $bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



    $bullet$ If $n<m$ then the limit is $0$.



    $bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



    Note that this method is only for limits to infinity






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
      The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



      For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



      Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
      {text{if};;limlimits_{x to a} gleft( x right) ne 0}$$       to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



      To Sum It Up



      Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

      Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



      $bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



      $bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



      $bullet$ If $n<m$ then the limit is $0$.



      $bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



      Note that this method is only for limits to infinity






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
        The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



        For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



        Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
        {text{if};;limlimits_{x to a} gleft( x right) ne 0}$$         to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



        To Sum It Up



        Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

        Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



        $bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



        $bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



        $bullet$ If $n<m$ then the limit is $0$.



        $bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



        Note that this method is only for limits to infinity






        share|cite|improve this answer









        $endgroup$



        The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials.
        The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $pm infty$.



        For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like.



        Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${limlimits_{x to a} frac{{fleft( x right)}}{{gleft( x right)}} }={ frac{{limlimits_{x to a} fleft( x right)}}{{limlimits_{x to a} gleft( x right)}},;;;}kern-0.3pt
        {text{if};;limlimits_{x to a} gleft( x right) ne 0}$$         to quickly get the answer by considering $$lim_{xto infty} frac{x^3}{x^5} = lim_{x to infty} frac{1}{x^2} = frac{1}{infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.



        To Sum It Up



        Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+cdots+b_{1}x+b_{0}.$$

        Then the value of $$lim_{xrightarrowinfty}frac{f(x)}{g(x)}$$ can be determined as follows:



        $bullet$ If $n>m$ and the sign of $a_n$ is the same as that of $b_m$ (i.e. $frac{a_{n}}{b_{m}}$ is positive) then the limit is $infty$



        $bullet$ If $n>m$ and the sign of $a_n$ is different from that of $b_m$, (i.e. $frac{a_{n}}{b_{m}}$ is negative) then the limit is $-infty$



        $bullet$ If $n<m$ then the limit is $0$.



        $bullet$ If $n=m$ then the limit is the value of $frac{a_{n}}{b_{m}}$.



        Note that this method is only for limits to infinity







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 11:45









        E.NoleE.Nole

        139114




        139114






























            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%2f2296069%2flimit-of-rational-function-solving%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

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more