Evaluate $ lim_{xto 0}|frac{5^x - 5^{-x}}{5^x-1}| $ without using L'Hospital's rule.












0












$begingroup$


$$
lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
$$



I know the limit is equal to 2. But I am not allowed to use L'Hospital.
How can I evaluate the limit without L'Hospital?






calculus limits limits-without-lhopital







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    $$
    lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
    $$



    I know the limit is equal to 2. But I am not allowed to use L'Hospital.
    How can I evaluate the limit without L'Hospital?






    calculus limits limits-without-lhopital







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $$
      lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
      $$



      I know the limit is equal to 2. But I am not allowed to use L'Hospital.
      How can I evaluate the limit without L'Hospital?






      calculus limits limits-without-lhopital







      share|cite|improve this question











      $endgroup$




      $$
      lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
      $$



      I know the limit is equal to 2. But I am not allowed to use L'Hospital.
      How can I evaluate the limit without L'Hospital?






      calculus limits limits-without-lhopital




      calculus limits limits-without-lhopital






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 30 '18 at 20:57









      Martin Sleziak

      44.7k9117272




      44.7k9117272










      asked Nov 30 '18 at 12:25









      nofnof

      32




      32






















          4 Answers
          4






          active

          oldest

          votes


















          8












          $begingroup$

          Hint:



          $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Set $y= 5^x$ and consider $ylongrightarrow 1$
            $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              HINT



              For positive $x$ we have:
              $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
              As $xrightarrow 0$ this approaches to $1+1=2$.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $






                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%2f3020035%2fevaluate-lim-x-to-0-frac5x-5-x5x-1-without-using-lhospital%23new-answer', 'question_page');
                  }
                  );

                  Post as a guest















                  Required, but never shown

























                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  8












                  $begingroup$

                  Hint:



                  $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$






                  share|cite|improve this answer









                  $endgroup$


















                    8












                    $begingroup$

                    Hint:



                    $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$






                    share|cite|improve this answer









                    $endgroup$
















                      8












                      8








                      8





                      $begingroup$

                      Hint:



                      $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint:



                      $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 30 '18 at 12:33









                      KM101KM101

                      5,9431523




                      5,9431523























                          2












                          $begingroup$

                          Set $y= 5^x$ and consider $ylongrightarrow 1$
                          $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Set $y= 5^x$ and consider $ylongrightarrow 1$
                            $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Set $y= 5^x$ and consider $ylongrightarrow 1$
                              $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$






                              share|cite|improve this answer









                              $endgroup$



                              Set $y= 5^x$ and consider $ylongrightarrow 1$
                              $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 30 '18 at 12:32









                              trancelocationtrancelocation

                              9,8101722




                              9,8101722























                                  2












                                  $begingroup$

                                  HINT



                                  For positive $x$ we have:
                                  $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
                                  As $xrightarrow 0$ this approaches to $1+1=2$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    HINT



                                    For positive $x$ we have:
                                    $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
                                    As $xrightarrow 0$ this approaches to $1+1=2$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      HINT



                                      For positive $x$ we have:
                                      $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
                                      As $xrightarrow 0$ this approaches to $1+1=2$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      HINT



                                      For positive $x$ we have:
                                      $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
                                      As $xrightarrow 0$ this approaches to $1+1=2$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 30 '18 at 12:52









                                      A. PongráczA. Pongrácz

                                      5,9831929




                                      5,9831929























                                          0












                                          $begingroup$

                                          Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                                          $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                                            $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                                              $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $






                                              share|cite|improve this answer









                                              $endgroup$



                                              Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                                              $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Nov 30 '18 at 12:36









                                              Shubham JohriShubham Johri

                                              4,785717




                                              4,785717






























                                                  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%2f3020035%2fevaluate-lim-x-to-0-frac5x-5-x5x-1-without-using-lhospital%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