Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$












2












$begingroup$


Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$



I did: $lim_{nto infty}(1+frac{1}{frac{n+1}{n-1}})^{frac{n+1}{n-1}}=e$.



Why is this incorrect?



Thank you for your help.






limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: What is the limit of $(n+1)/(n-1)$ ?
    $endgroup$
    – Martin R
    Jan 23 at 12:40






  • 2




    $begingroup$
    Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
    $endgroup$
    – xbh
    Jan 23 at 12:41










  • $begingroup$
    limit of $(n+1)/(n-1)=1$, thank you!
    $endgroup$
    – J.Doe
    Jan 23 at 12:41


















2












$begingroup$


Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$



I did: $lim_{nto infty}(1+frac{1}{frac{n+1}{n-1}})^{frac{n+1}{n-1}}=e$.



Why is this incorrect?



Thank you for your help.






limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: What is the limit of $(n+1)/(n-1)$ ?
    $endgroup$
    – Martin R
    Jan 23 at 12:40






  • 2




    $begingroup$
    Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
    $endgroup$
    – xbh
    Jan 23 at 12:41










  • $begingroup$
    limit of $(n+1)/(n-1)=1$, thank you!
    $endgroup$
    – J.Doe
    Jan 23 at 12:41
















2












2








2





$begingroup$


Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$



I did: $lim_{nto infty}(1+frac{1}{frac{n+1}{n-1}})^{frac{n+1}{n-1}}=e$.



Why is this incorrect?



Thank you for your help.






limits







share|cite|improve this question











$endgroup$




Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$



I did: $lim_{nto infty}(1+frac{1}{frac{n+1}{n-1}})^{frac{n+1}{n-1}}=e$.



Why is this incorrect?



Thank you for your help.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 12:52









amWhy

1




1










asked Jan 23 at 12:37









J.DoeJ.Doe

899




899








  • 2




    $begingroup$
    Hint: What is the limit of $(n+1)/(n-1)$ ?
    $endgroup$
    – Martin R
    Jan 23 at 12:40






  • 2




    $begingroup$
    Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
    $endgroup$
    – xbh
    Jan 23 at 12:41










  • $begingroup$
    limit of $(n+1)/(n-1)=1$, thank you!
    $endgroup$
    – J.Doe
    Jan 23 at 12:41
















  • 2




    $begingroup$
    Hint: What is the limit of $(n+1)/(n-1)$ ?
    $endgroup$
    – Martin R
    Jan 23 at 12:40






  • 2




    $begingroup$
    Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
    $endgroup$
    – xbh
    Jan 23 at 12:41










  • $begingroup$
    limit of $(n+1)/(n-1)=1$, thank you!
    $endgroup$
    – J.Doe
    Jan 23 at 12:41










2




2




$begingroup$
Hint: What is the limit of $(n+1)/(n-1)$ ?
$endgroup$
– Martin R
Jan 23 at 12:40




$begingroup$
Hint: What is the limit of $(n+1)/(n-1)$ ?
$endgroup$
– Martin R
Jan 23 at 12:40




2




2




$begingroup$
Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
$endgroup$
– xbh
Jan 23 at 12:41




$begingroup$
Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
$endgroup$
– xbh
Jan 23 at 12:41












$begingroup$
limit of $(n+1)/(n-1)=1$, thank you!
$endgroup$
– J.Doe
Jan 23 at 12:41






$begingroup$
limit of $(n+1)/(n-1)=1$, thank you!
$endgroup$
– J.Doe
Jan 23 at 12:41












2 Answers
2






active

oldest

votes


















6












$begingroup$

I hope you solved the problem. Here is a fact that might help you.



Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
    But, you can rewrite it as
    $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
    so, the limit is
    $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$






    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%2f3084414%2fcompute-lim-n-to-infty1-fracn-1n1-fracn1n-1%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









      6












      $begingroup$

      I hope you solved the problem. Here is a fact that might help you.



      Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        I hope you solved the problem. Here is a fact that might help you.



        Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          I hope you solved the problem. Here is a fact that might help you.



          Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$






          share|cite|improve this answer









          $endgroup$



          I hope you solved the problem. Here is a fact that might help you.



          Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 12:50









          Sujit BhattacharyyaSujit Bhattacharyya

          1,372519




          1,372519























              0












              $begingroup$

              Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
              But, you can rewrite it as
              $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
              so, the limit is
              $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
                But, you can rewrite it as
                $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
                so, the limit is
                $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
                  But, you can rewrite it as
                  $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
                  so, the limit is
                  $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$






                  share|cite|improve this answer









                  $endgroup$



                  Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
                  But, you can rewrite it as
                  $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
                  so, the limit is
                  $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 13:09









                  CarlIOCarlIO

                  957




                  957






























                      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%2f3084414%2fcompute-lim-n-to-infty1-fracn-1n1-fracn1n-1%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

                      Mario Kart Wii

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                      Read more

                      What does “Dominus providebit” mean?

                      Image
                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                      Read more

                      Antonio Litta Visconti Arese

                      Image
                      Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                      Read more